Question

A 97.5-kg man pushes a crate 6.5 m up along a ramp that makes an angle of 19.5° with the horizontal. He exerts a force of 490 N on the crate parallel to the ramp and moves it at a constant speed.

Answer 1

The work done by the man in pushing the crate up the ramp is calculated to be approximately 3,797 J.

Step-by-step explanation:

When an object is moved at a constant speed, the net force acting on it is zero. In this case, the force exerted by the man (Fman) is balanced by the frictional force (ffriction) and the component of the weight of the crate (W∥) parallel to the ramp.

Let's break down the forces along the incline:

Force exerted by the man (Fman ): This force is parallel to the ramp, and its magnitude is given as 490N.

Weight of the crate (W): The weight of the crate can be decomposed into two components:

W∥ is the component parallel to the ramp.

W⊥ is the component perpendicular to the ramp.

The weight W can be calculated as W=m⋅g, where m is the mass of the crate (97.5kg) and g is the acceleration due to gravity (9.8 m/s2).

W∥ =W⋅sin(θ)

W⊥ =W⋅cos(θ)

Frictional force (ffriction): The frictional force acts parallel to the ramp and opposes the motion. At constant speed, the frictional force is equal in magnitude and opposite in direction to the applied force.

ffriction=Fman

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Now, the equation for forces along the incline can be written as:

Fman+ffriction +W∥ =0

Substitute the known values into the equation and solve for W∥

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490N+ffriction +W∥ =0

Since the crate is moving at a constant speed, ffriction can be directly replaced by Fman(equal in magnitude but opposite in direction):

490N−490N+W∥=0

Now, solve for W∥ , which represents the component of the weight parallel to the ramp. This force represents the force that the man needs to overcome to move the crate at a constant speed up the ramp.