Question

A 25kg block is pushed a distance 5m along a level floor at constant speed by force at an angle of 30° below the horizontal. The coefficient of friction between the block and the floor is 0.4 how much work is done?​

Answer 1

The coefficient of friction between the block and the floor is 0.4 means that the force of friction is 40 percent of the force of the push. So, the force of friction is 30 kg. The force of the push is 25 kg. So, the work done is the force of the push times the distance the block moves, which is : 25 kg x 5 m. This means that 125 joules of work are done.

Answer 2

To find the work done, we need to calculate the force applied to the block and the distance it moves.

The force applied to the block can be resolved into two components: one parallel to the floor and one perpendicular to the floor. The parallel component of the force is responsible for pushing the block forward, while the perpendicular component does not contribute to the work done.

The parallel component of the force is:

F_parallel = F * cos(30°) = F * √3/2

where F is the magnitude of the force applied.

The force of friction opposing the motion is:

F_friction = μ * F_norm

where μ is the coefficient of friction and F_norm is the normal force acting on the block, which is equal to the weight of the block since it is on a level floor:

F_norm = m * g = 25 kg * 9.81 m/s^2 = 245.25 N

where g is the acceleration due to gravity.

So the force of friction is:

F_friction = 0.4 * 245.25 N = 98.1 N

Since the block is moving at constant speed, the force applied must be equal and opposite to the force of friction:

F_parallel = F_friction

F * √3/2 = 98.1 N

F = 56.6 N

The work done by the force applied is:

W = F_parallel * d = 56.6 N * 5 m = 283 J

Therefore, the work done by the force applied is 283 J