Question

A woman pushes a m = 3.00 kg block a distance d = 2.80 m along the floor by a constant force of magnitude F = 16.0 N directed at an angle = 27.0° below the horizontal as shown in the figure. Assume the floor is frictionless. (Enter your answers in joules.)

(a)

Determine the work done on the block by the applied force (the force on the block exerted by the woman).



(b)

Determine the work done on the block by the normal force exerted by the floor.



(c)

Determine the work done on the block by the gravitational force.



(d)

Determine the work done by the net force on the block.

Answer 1

The work done on the block is 37.588 J by the applied force, 0 J by the normal force, -82.656 J by the gravitational force, and the net work done is -45.068 J.

Step-by-step explanation:

(a) The work done on the block by the applied force can be calculated using the formula:

Work = Force * Distance * Cosine(theta)

where Force is the magnitude of the force, Distance is the distance moved by the block, and theta is the angle between the force and the displacement. Plugging in the given values, we get:

Work = 16.0 N * 2.80 m * Cos(27.0°) = 37.588 J

(b) The work done on the block by the normal force exerted by the floor is zero. This is because the normal force is perpendicular to the displacement of the block.

(c) The work done on the block by the gravitational force can be calculated using the formula:

Work = Force * Distance * Cosine(theta)

where Force is the weight of the block, Distance is the distance moved by the block, and theta is the angle between the force and the displacement. Plugging in the given values, we get:

Work = (3.00 kg * 9.8 m/s^2) * 2.80 m * Cos(180°) = -82.656 J

(d) The work done by the net force on the block can be calculated by adding the work done by the applied force, the normal force, and the gravitational force. So:

Net Work = Work by Applied Force + Work by Normal Force + Work by Gravitational Force

Net Work = 37.588 J + 0 J + (-82.656 J) = -45.068 J