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Q10:
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution :
Step 1: Theoretical Foundation
According to Newton's First Law of Motion, an object maintains its state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force [Law of Inertia].
The problem states that the cabinet is being moved at a constant velocity ($v = \text{constant}$). In physics, constant velocity implies zero acceleration ($a = 0$), because acceleration is defined as the rate of change of velocity ($a = \frac{dv}{dt}$).
Step 2: Applying Newton's Second Law of Motion
Newton's Second Law states that the net force ($\sum F$) acting on an object is equal to the product of its mass ($m$) and its acceleration ($a$):
$$\sum F = m \times a$$
Since the cabinet moves at a constant velocity, the acceleration $a = 0 \text{ m/s}^2$. Substituting this into the equation:
$$\sum F = m \times 0 = 0 \text{ N}$$
This confirms that the net force acting on the cabinet must be zero for it to maintain a constant velocity [Equilibrium Condition].
Step 3: Analyzing the Force Balance
The horizontal forces acting on the cabinet are the applied force ($F_{\text{applied}}$) and the frictional force ($F_{\text{friction}}$) acting in the opposite direction. The vector sum of these forces must equal the net force:
$$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$$
Given $F_{\text{net}} = 0 \text{ N}$ and $F_{\text{applied}} = 200 \text{ N}$, we substitute the values:
$$0 = 200 \text{ N} - F_{\text{friction}}$$
$$F_{\text{friction}} = 200 \text{ N}$$
Vector Representation of Forces
Data Summary
| Variable | Value |
|---|---|
| Applied Force ($F_a$) | $200 \text{ N}$ |
| Acceleration ($a$) | $0 \text{ m/s}^2$ |
| Net Force ($F_{net}$) | $0 \text{ N}$ |
| Frictional Force ($F_f$) | $200 \text{ N}$ |
Final Solution: The frictional force exerted on the cabinet is 200 N, which perfectly opposes the applied force to maintain dynamic equilibrium.
More Questions from Class 9 Science Force and Laws of Motion Exercises
- Q1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on
- Q2: When a carpet is beaten with a stick, dust comes out of it, Explain.
- Q3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?
- Q4: A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
- Q5: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
- Q6: A stone of 1 kg is thrown with a velocity of 20 $m s^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
- Q7(a): A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force
- Q7(b): A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (b) the acceleration of the train.
- Q8: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 $m s^{-2}$?
- Q9: What is the momentum of an object of mass $m$, moving with a velocity $v$?
- Q11: According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
- Q12: A hockey ball of mass 200 g travelling at 10 $m s^{-1}$ is struck by a hockey stick so as to return it along its original path with a velocity at 5 $m s^{-1}$. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
- Q13: A bullet of mass 10 g travelling horizontally with a velocity of 150 $m s^{-1}$ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
- Q14: An object of mass 1 kg travelling in a straight line with a velocity of 10 $m s^{-1}$ collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
- Q15: An object of mass 100 kg is accelerated uniformly from a velocity of 5 $m s^{-1}$ to 8 $m s^{-1}$ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
- Q16: Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
- Q17: How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 $m s^{-2}$.
CBSE Solutions for Class 9 Science Force and Laws of Motion
Chapters in CBSE - Class 9 Science
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