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A block of mass 5.0 kg is pushed up a 30 degree incline by a constant horizontal force of...

Question:

A block of mass 5.0 kg is pushed up a 30{eq}^{\circ} {/eq} incline by a constant horizontal force of magnitude 60.0 N. The block travels a distance of 0.25 m along the incline and is subject to a constant frictional force of 3.0 N.

a) Calculate the work done by the applied force.

b) Calculate the work done by the frictional force.

c) Calculate the work done by the gravitational force acting on the block.

d) Calculate the work done by the normal force acting on the block.

Force and Work:

Work is the product of force and distance. It is important to understand that the distance you consider is the distance in the direction of force.

Answer and Explanation:

Part(a)

Work done by applied force is equal to the product of applied force and distance i.e. {eq}60(0.25) = 15J {/eq}

Part(b)

Work done by friction is equal to the product of frictional force and distance i.e. {eq}3(0.25) = 0.75J {/eq}

Part(c)

Work done by gravitational force is equal to {eq}mgh = 5(9.81)(0.25)sin 30 = 6.13J {/eq}

Part(d)

Distance moved in the direction of normal force is {eq}0.25\tan 30 = 0.144m {/eq}

The work done is the product of normal force and distance {eq}= 5\cos 30 \times 0.144 = 0.624J {/eq}


Learn more about this topic:

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Newton's Second Law: Physics Lab

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