A block of mass 2 kg, moving with the initial speed of 3 m/s comes to rest on a rough horizontal surface after travelling a distance of 3 m. The magnitude of the frictional force is:
- (a)9 N
- (b)3 N
- (c)18 N
- (d)1 N
Correct — B, 3 N. The block's entire kinetic energy is used up doing work against friction over the 3 m it travels before stopping. Its initial kinetic energy is (1/2)mv^2 = (1/2)(2 kg)(3 m/s)^2 = 9 J. Setting this equal to the work done by friction, f x d = 9 J with d = 3 m, gives f = 9 / 3 = 3 N. The same answer comes from mechanics: using v^2 = u^2 - 2as with v = 0, u = 3 m/s and s = 3 m gives a deceleration of 1.5 m/s^2, and by Newton's second law the frictional force is f = ma = 2 x 1.5 = 3 N.
- (a)9 N — 9 is the value of the initial kinetic energy in joules, not the force in newtons. The energy must still be divided by the 3 m stopping distance to get the force, which gives 3 N.
- (c)18 N — 18 follows from forgetting the factor of one-half and using mv^2 = 2 x 9 = 18 J as the energy. The correct kinetic energy is (1/2)mv^2 = 9 J, so the force is 3 N, not 18 N.
- (d)1 N — This underestimates the force. The energy balance (1/2)mv^2 = f x d fixes f at 3 N; 1 N would leave the block still moving well past 3 m.
The work-energy theorem states that the work done by the net force on a body equals the change in its kinetic energy. When a moving block is brought to rest by friction on a level surface, friction is the only force doing work along the motion, so the magnitude of that work equals the kinetic energy the block loses.
The tempting error is to read off the kinetic energy (9 J) and quote it as the force, or to drop the one-half in (1/2)mv^2. The friction force is found by dividing the dissipated energy by the distance over which it acts, so the joules must be turned into newtons using the 3 m path length.
- Work-energy theorem: work done by the net force equals the change in kinetic energy.
- Kinetic energy of the block = (1/2)(2 kg)(3 m/s)^2 = 9 J.
- Work done against friction = force x distance = f x 3 m, set equal to 9 J gives f = 3 N.
- Cross-check by kinematics: v^2 = u^2 - 2as gives a = 1.5 m/s^2, so f = ma = 3 N.
- Quoting the kinetic energy in joules as if it were the force in newtons, without dividing by the distance.
- Forgetting the factor of one-half in (1/2)mv^2, which doubles the answer to 18 N.
A numerical work-energy problem: equate the initial kinetic energy to the work done against friction over the stopping distance to get the force.
No directly related past PYQ was found.
- practice — not a real PYQ
A 1 kg block moving at 4 m/s is brought to rest by friction after sliding 2 m on a rough floor. The magnitude of the frictional force is
- (a)2 N
- (b)4 N
- (c)8 N
- (d)16 N
Answer(b) 4 N — kinetic energy (1/2)(1)(4^2) = 8 J equals f x 2 m, so f = 4 N.
- practice — not a real PYQ
The work-energy theorem states that the work done by the net force acting on a body equals the change in its
- (a)momentum
- (b)kinetic energy
- (c)potential energy
- (d)acceleration
Answer(b) kinetic energy — the net work done on a body equals the change in its kinetic energy.