High School

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A 98 kg block is pulled at a constant speed of [tex]9.2 \, \text{m/s}[/tex] across a horizontal floor by an applied force of [tex]126 \, \text{N}[/tex] directed [tex]18^\circ[/tex] above the horizontal.

What is the rate at which the force does work on the block?

Answer :

To calculate the rate at which the applied force does work on the 98kg block, compute the power by finding the horizontal component of the force and multiplying it by the block's velocity.

The question asks for the rate at which the force does work on a block being pulled across a horizontal floor. The rate of doing work in physics is known as power. To calculate the power, one must find the component of the force in the direction of motion, which is obtained by multiplying the total force by the cosine of the angle of the force to the horizontal. Given a force of 126N at an 18° angle above the horizontal and a block speed of 9.2 m/s, the horizontal force component can be calculated as follows: Fhorizontal = 126N imes cos(18°).

The power done by the force is then the product of this horizontal component and the velocity of the block: Power = Fhorizontal imes velocity. Substituting the known values, we compute the power transferred to the block, representing the rate at which the force does work on the block.

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