By Newton's first law, a body at rest or moving with constant velocity has zero acceleration. Since $F_{net}=ma$, the net force is $0$ in each case. Any individual forces present balance one another.
In all five cases, the net force is zero.
With air resistance ignored, the only force on the pebble after it is thrown is its weight. $W=mg=0.05\times 10=0.5\,\text{N}$ downward. This remains true while the pebble rises, falls, and at the highest point.
The net force is $0.5\,\text{N}$ vertically downward in all cases. The answer does not change for a 45° projection if air resistance is ignored.
For (a), (b), and (c), once the stone is released and air resistance is neglected, the only force on it is gravity: $mg=0.1\times 10=1\,\text{N}$ downward. In (d), the floor supports the stone vertically, so weight and normal reaction cancel. To keep the stone at rest relative to the accelerating train, a horizontal frictional force $ma=0.1\times 1=0.1\,\text{N}$ acts forward.
(a) $1\,\text{N}$ downward. (b) $1\,\text{N}$ downward. (c) $1\,\text{N}$ downward. (d) $0.1\,\text{N}$ horizontally in the direction of the train's acceleration.
- i. T
- ii. T - mv^2/l
- iii. T + mv^2/l
- iv. 0
On a smooth horizontal table, the only horizontal force towards the centre is the string tension. Therefore the net centripetal force is $T$, and for circular motion $T=mv^2/l$.
(i) $T$.
The retardation is $a=F/m=50/20=2.5\,\text{m s}^{-2}$. With final speed zero, $0=15-2.5t$, so $t=15/2.5=6\,\text{s}$.
$6\,\text{s}$.
$a=(v-u)/t=(3.5-2.0)/25=0.06\,\text{m s}^{-2}$. Hence $F=ma=3.0\times 0.06=0.18\,\text{N}$. Since the speed increases without changing direction, the force acts along the motion.
$0.18\,\text{N}$ in the direction of motion.
The resultant force is $F=\sqrt{8^2+6^2}=10\,\text{N}$. Thus $a=F/m=10/5=2\,\text{m s}^{-2}$. The angle with the 8 N force is $\tan^{-1}(6/8)=36.9^\circ$.
$2\,\text{m s}^{-2}$, at $36.9^\circ$ to the 8 N force towards the 6 N force.
$36\,\text{km h}^{-1}=10\,\text{m s}^{-1}$. The acceleration is $a=(0-10)/4=-2.5\,\text{m s}^{-2}$. Total mass $=400+65=465\,\text{kg}$. Therefore $F=ma=465\times (-2.5)=-1162.5\,\text{N}$, so the retarding force has magnitude $1.16\times 10^3\,\text{N}$.
$1.16\times 10^3\,\text{N}$ opposite to the motion.
For upward acceleration, $T-mg=ma$. Hence $T=m(g+a)=20000(10+5)=3.0\times 10^5\,\text{N}$.
$3.0\times 10^5\,\text{N}$ upward.
The acceleration during the 30 s force interval is $a=-8.0/0.40=-20\,\text{m s}^{-2}$. Before $t=0$, motion is uniform, so $x=10t$ and $x(-5)=-50\,\text{m}$. At $t=25\,\text{s}$, $x=ut+\frac12at^2=10(25)-10(25)^2=-6000\,\text{m}$. At $t=30\,\text{s}$, $x=10(30)-10(30)^2=-8700\,\text{m}$ and $v=10-20(30)=-590\,\text{m s}^{-1}$. From 30 s to 100 s the force is absent, so $x(100)=-8700+(-590)(70)=-50000\,\text{m}$.
Taking north as positive: $x(-5\,\text{s})=-50\,\text{m}$, $x(25\,\text{s})=-6000\,\text{m}$, and $x(100\,\text{s})=-50000\,\text{m}$.
At $t=10\,\text{s}$, the truck's speed is $v_x=at=2.0\times 10=20\,\text{m s}^{-1}$. The dropped stone keeps this horizontal velocity. After 1 s of fall, $v_y=-gt=-10\,\text{m s}^{-1}$. Thus $\vec v=20\hat{i}-10\hat{j}\,\text{m s}^{-1}$ and $|\vec v|=\sqrt{20^2+10^2}=22.4\,\text{m s}^{-1}$. Its acceleration is only due to gravity.
At $t=11\,\text{s}$, velocity is $20\hat{i}-10\hat{j}\,\text{m s}^{-1}$, with speed $22.4\,\text{m s}^{-1}$ at $26.6^\circ$ below the horizontal. Acceleration is $10\,\text{m s}^{-2}$ downward.
At an extreme position the bob's instantaneous speed is zero, so after the string is cut it starts from rest and falls vertically under gravity. At the mean position, the bob has horizontal tangential velocity; after the cut, gravity acts downward, so the motion is projectile motion along a parabola.
(a) It falls vertically downward. (b) It follows a parabolic projectile path tangent to the swing at the mean position.
The scale reads the normal reaction, usually calibrated as an equivalent mass $N/g$. (a) Uniform speed means $a=0$, so $N=mg=700\,\text{N}$. (b) For downward acceleration, $N=m(g-a)=70(10-5)=350\,\text{N}$. (c) For upward acceleration, $N=m(g+a)=70(10+5)=1050\,\text{N}$. (d) In free fall, $a=g$ downward, so $N=m(g-g)=0$.
(a) $70\,\text{kg}$ reading ($700\,\text{N}$ normal force). (b) $35\,\text{kg}$ reading ($350\,\text{N}$). (c) $105\,\text{kg}$ reading ($1050\,\text{N}$). (d) Zero.
From the graph, velocity is $0$ before $t=0$, then $v=(3-0)/(4-0)=0.75\,\text{m s}^{-1}$ for $0<t<4\,\text{s}$, and $0$ again after $t=4\,\text{s}$. In each interval velocity is constant, so acceleration and force are zero. Impulse equals change in momentum: at $t=0$, $J=m(0.75-0)=4\times0.75=3\,\text{N s}$; at $t=4\,\text{s}$, $J=m(0-0.75)=-3\,\text{N s}$.
(a) The force is zero for $t<0$, $0<t<4\,\text{s}$, and $t>4\,\text{s}$. (b) Impulse at $t=0$ is $+3\,\text{N s}$; impulse at $t=4\,\text{s}$ is $-3\,\text{N s}$.
The common acceleration is $a=600/(10+20)=20\,\text{m s}^{-2}$. If A is pulled, the tension accelerates the 20 kg body, so $T=20\times20=400\,\text{N}$. If B is pulled, the tension accelerates the 10 kg body, so $T=10\times20=200\,\text{N}$.
If the 600 N force is applied to the 10 kg body A, $T=400\,\text{N}$. If it is applied to the 20 kg body B, $T=200\,\text{N}$.
For an Atwood machine, $a=\frac{(12-8)g}{12+8}=\frac{4\times10}{20}=2\,\text{m s}^{-2}$. For the 8 kg mass moving upward, $T-8g=8a$, so $T=8(10+2)=96\,\text{N}$.
Acceleration is $2\,\text{m s}^{-2}$, with the 12 kg mass moving downward and the 8 kg mass moving upward. Tension is $96\,\text{N}$.
The initial momentum of the nucleus is zero. With no external force during disintegration, momentum is conserved: $\vec p_1+\vec p_2=0$. Therefore $\vec p_1=-\vec p_2$, which means the two products move in opposite directions.
The two product nuclei must have equal and opposite momenta, so they move in opposite directions.
For either ball, the initial momentum is $mu=0.05\times6=0.30\,\text{kg m s}^{-1}$ in its initial direction, and after rebounding it is $-0.30\,\text{kg m s}^{-1}$. Thus $J=\Delta p=-0.30-0.30=-0.60\,\text{N s}$ for one ball. The other receives an equal impulse in the opposite direction.
Each ball receives an impulse of magnitude $0.60\,\text{N s}$, opposite to its initial direction of motion.
Initial momentum is zero. By conservation of momentum, $m_s v_s + M_g v_g=0$. Hence $v_g=-(0.020\times80)/100=-0.016\,\text{m s}^{-1}$. The negative sign shows recoil opposite to the shell.
$0.016\,\text{m s}^{-1}$ opposite to the shell's motion.
$54\,\text{km h}^{-1}=15\,\text{m s}^{-1}$. Since the speed is unchanged and the velocity direction changes by $45^\circ$, $|\Delta v|=2v\sin(45^\circ/2)=2(15)\sin22.5^\circ=11.48\,\text{m s}^{-1}$. Therefore $J=m|\Delta v|=0.15\times11.48=1.72\,\text{N s}$.
The impulse has magnitude about $1.7\,\text{N s}$, directed along the change in the ball's velocity.
The angular speed is $\omega=40(2\pi)/60=4\pi/3\,\text{rad s}^{-1}$. Tension supplies centripetal force: $T=m\omega^2r=0.25(4\pi/3)^2(1.5)\approx6.6\,\text{N}$. For maximum tension, $T_{max}=mv^2/r$, so $v_{max}=\sqrt{T_{max}r/m}=\sqrt{200\times1.5/0.25}=34.6\,\text{m s}^{-1}$.
Tension is about $6.6\,\text{N}$. Maximum speed is $34.6\,\text{m s}^{-1}$.
- a. the stone moves radially outwards
- b. the stone flies off tangentially from the instant the string breaks
- c. the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?
At the instant the string breaks, the centripetal force disappears. The stone continues with its instantaneous velocity, which is tangential to the circular path.
(b) the stone flies off tangentially from the instant the string breaks.
A horse-cart system needs an external horizontal force from the ground; in empty space the horse cannot push against the ground to get frictional reaction. In a stopping bus, the lower body slows with the bus but the upper body tends to keep its forward motion. For a lawn mower, friction is proportional to normal reaction; pulling reduces the normal reaction, pushing increases it. For a catch, impulse equals change in momentum, $J=F_{avg}\Delta t$; increasing stopping time lowers the average force.
(a) There is no ground friction in empty space to provide the external forward force. (b) Due to inertia, passengers tend to continue moving forward. (c) Pulling gives an upward component that reduces the normal reaction and friction, while pushing gives a downward component that increases them. (d) Moving the hands backward increases the stopping time, reducing the average force on the hands.