A 4.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 29.0° with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s
If the coefficient of kinetic friction between the block and the wall is 0.30, find
a) the work done by the force on the block.
Answer in units of J.
To find the work done by the force on the block, we can break it down into two components: the force parallel to the displacement, and the force perpendicular to the displacement.
The force parallel to the displacement can be found by multiplying the applied force by the cosine of the angle between the force and the displacement. In this case, the magnitude of the applied force can be found using the formula F = ma, where m is the mass of the block and a is the acceleration, which is zero since the block is moving at a constant velocity. Therefore, F = 0, and the force parallel to the displacement is also zero.
The force perpendicular to the displacement can be found by multiplying the applied force by the sine of the angle between the force and the displacement. In this case, the perpendicular force is equal to the weight of the block. The weight can be found using the formula W = mg, where m is the mass of the block and g is the acceleration due to gravity. Therefore, W = 4.0 kg * 9.81 m/s^2 = 39.24 N.
The work done by the perpendicular force is calculated by multiplying the force by the displacement. In this case, the displacement is in the vertical direction, so the work done by the perpendicular force is W * d = 39.24 N * 2.0 m = 78.48 J.
Therefore, the work done by the force on the block is 78.48 J.