Question

A 19.8 kg block is dragged over a rough, horizontal surface by a constant force of 174 N
acting at an angle of 25.8 degrees above the horizontal. The block is displaced 63.9 m, and the coefficient of kinetic friction is 0.123.

Part A: Find the work done by the 174 N force. The
acceleration of gravity is 9.8 m/s^2
Answer in units of J.

Part B: Find the magnitude of the work done by the
force of friction.
Answer in units of J.

Answer 1

→ The answers are listed below:

Part A) Work = 10010.28 J

Part B) Work = 929.879 J

To answer part A we should apply the following equation of Work:

$\Large \text { Work = Force \times Distance \times cos(\Theta) }$

Therefore:

$\Large \text { Work = 174 \times 63.9 \times cos(25.8) }\\\\\Large \text { Work = 10010.28 J}$

Now, at part B, the magnitude of the work done by the force of friction can also be determinated using the same equation, but before we have to find the force. On the free body diagram, we can apply the second Law of Newton to find the value of 'N'. Remember that in 'y' axis the block is not moving, so:

$\Large \text { N + Fy = P }\\\\\Large \text { N = P - Fy }\\\\\Large \text { N = m \times g - 174 \times sin(\Theta) }\\\\\Large \text { N = 19.8 \times 9.8 - 174 \times sin(25.8) }\\\\\Large \text { N = 118.309 Newtons}$

Now, keep in mind that the friction force can be determinated by

$\Large \text { Fa = N \times \mu }$

μ is the coefficient of kinetic friction.

$\Large \text { Fa = 86.362 \times 0.123 = 14.5521 Newtons}$

Finally, we can determine the work ()

$\Large \text { Fa = 14.5521 \times 63.9 = 929.879 J}$

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