A 40 kg block is pushed 2.0 c at a conetant velocity up a vertical wall by a constant force applied at an angle of 200° with the horizon-tal, as shown in the figure.
The acccloration of gravity is 9.81 m/
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L a Hooke'e-law
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1 kg
Drawing not to scale.
If the coefficient of kinetic briction between the block and the wall is 0.30, find
a) the work done by the force on the block.
Answer in units of 1.
To find the work done by the force on the block, we need to calculate the displacement of the block in the direction of the force applied.
Given:
Mass of the block (m) = 40 kg
Force applied (F) = ? (not given)
Distance moved (s) = 2.0 m
Angle between force and horizontal (θ) = 200°
Coefficient of kinetic friction (μ) = 0.30
Acceleration due to gravity (g) = 9.81 m/s^2
To find the force applied (F), we need to resolve it into its horizontal and vertical components:
Horizontal component of force (F_h) = F * cos(θ)
Vertical component of force (F_v) = F * sin(θ)
Since the block is moving at a constant velocity, the net force in the vertical direction must be zero:
F_v - mg = 0
F_v = mg
Substituting the value of mass and acceleration due to gravity:
F * sin(θ) = (40 kg) * (9.81 m/s^2)
F * sin(200°) = 392.4 N
Therefore, the force applied (F) is:
F = 392.4 N / sin(200°)
F ≈ -942.04 N (The negative sign indicates that the force is pointing downwards)
Now, we can calculate the work done by the force on the block:
Work done (W) = F * s * cos(θ)
W = (-942.04 N) * (2.0 m) * cos(200°)
Using a calculator:
W ≈ -470 J
The work done by the force on the block is approximately -470 joules.