2.0 kg block is pushed 1.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 28.0° with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s?
Your
28°
2 kg
Drawing not to scale.
If the coefficient of kinetic friction between the block and the wall is 0.20, find
a) the work done by the force on the block.
Answer in units of J.
To calculate the work done by the force on the block, we need to find the component of the force that is parallel to the displacement of the block.
The force applied to the block can be broken down into two components: the force parallel to the displacement (F_parallel) and the force perpendicular to the displacement (F_perpendicular).
The force parallel to the displacement can be calculated using the formula:
F_parallel = F * cos(theta)
where F is the magnitude of the force and theta is the angle between the force and the horizontal (28.0°).
The magnitude of the force can be calculated using Newton's second law:
F = m * a
where m is the mass of the block (2.0 kg) and a is the acceleration (which is zero since the block is moving at a constant velocity).
Therefore, F = 2.0 kg * 0 m/s^2 = 0 N.
Now let's calculate the work done by the force:
Work = F_parallel * d
where d is the displacement (1.0 m).
Substituting the values we have:
Work = (0 N) * (1.0 m) = 0 J.
Therefore, the work done by the force on the block is 0 J.