Explain why
$(a)$ a horse cannot pull a cart and run in empty space,
$(b)$ passengers are thrown forward from their seats when a speeding bus stops suddenly,
$(c)$ it is easier to pull a lawn mower than to push it,
$(d)$ a cricketer moves his hands backwards while holding a catch.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) In order to pull a cart,a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. Therefore,a horse cannot pull a cart and run in empty space.
$(b)$ When a speeding bus stops suddenly,the lower portion of a passenger's body,which is in contact with the seat,suddenly comes to rest. However,the upper portion tends to remain in motion (as per the first law of motion). As a result,the passenger's upper body is thrown forward in the direction in which the bus was moving.
$(c)$ While pulling a lawn mower,a force at an angle $\theta$ is applied on it. The vertical component of this applied force $(F \sin \theta)$ acts upward,which reduces the effective weight of the mower $(mg - F \sin \theta)$. On the other hand,while pushing a lawn mower,the vertical component of the applied force acts in the direction of the weight of the mower,increasing the effective weight $(mg + F \sin \theta)$. Since the effective weight is lesser in the first case,pulling the lawn mower is easier than pushing it.
$(d)$ According to Newton's second law of motion,$F = ma = m \frac{\Delta v}{\Delta t}$. The impact force $(F)$ is inversely proportional to the impact time $(\Delta t)$,i.e.,$F \propto \frac{1}{\Delta t}$. By moving his hands backward,a cricketer increases the time of impact $(\Delta t)$,which results in a decrease in the stopping force,thereby preventing the hands from getting hurt.

Explore More

Similar Questions

$A$ block of mass $M = 100 \, kg$ is pulled by a force $T$ at an angle of $37^{\circ}$ with the horizontal. The coefficient of friction between the block and the ground is $\mu = 1/3$. $A$ person of mass $m = 25 \, kg$ is standing on a platform and pulls the block with the force $T$. Find the maximum upward acceleration $a_{\max}$ of the person such that the block just starts to move.

Difficult
View Solution

In the figure,two blocks $M$ and $m$ are tied together with an inextensible and light string. The mass $M$ is placed on a rough horizontal surface with coefficient of friction $\mu$ and the mass $m$ is hanging vertically against a smooth vertical wall. The pulley is frictionless. When the downward acceleration of the elevator becomes equal to $g$,then

Difficult
View Solution

$A$ boy of mass $4 \, kg$ is standing on a piece of wood having mass $5 \, kg$. If the coefficient of friction between the wood and the floor is $0.5$,the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ...... $N$. (Round off to the Nearest Integer) [Take $g = 10 \, m s^{-2}$]

From Newton's second law of motion,it can be inferred that

$A$ bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of $25 \, N$. The bird (mass $m = 0.5 \, kg$) flies upward in the cage with an acceleration of $2 \, m/s^2$. The spring balance will now record a weight of ........ $N$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo