A 2.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 30.0° with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s?
Your response...
PALETTE
30°
2 1
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 find the work done by the force on the block, we first need to determine the net force acting on the block.
The force pushing the block up the wall can be split into two components: the horizontal component and the vertical component.
The horizontal component of the force can be found by multiplying the magnitude of the force by the cosine of the angle: Fx = F * cos(30°).
The vertical component of the force can be found by multiplying the magnitude of the force by the sine of the angle: Fy = F * sin(30°).
Since the block is moving at a constant velocity, the net force must be zero. Therefore, the force of friction acting on the block must be equal in magnitude and opposite in direction to the component of the force pushing the block down the wall.
The force of friction can be found by multiplying the coefficient of kinetic friction (μ) by the normal force (mg), where m is the mass of the block and g is the acceleration due to gravity: f = μmg.
Setting the force of friction equal to the component of the force pushing the block down the wall gives: μmg = Fy.
We can rearrange this equation to solve for the force of friction: f = μmg = Fy.
The work done by the force on the block is given by the equation: work = force * distance.
Since the block is moving vertically, the distance traveled is given by the height of the wall, which is 2.0 m.
Substituting our values into the equation, we have: work = Fy * distance = μmg * distance.
Now we can solve for the work done by the force on the block:
work = (0.20) * (2.0 kg) * (9.81 m/s^2) * (2.0 m) = 7.85 J.
Therefore, the work done by the force on the block is 7.85 J.