123 Please show work last chegg expert got most answers wron

12.3:

Please show work, last chegg expert got most answers wrong.

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 95.0 kg, down a theta = 52.0 degree slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100. How much work is done by friction as the sled moves 30.0 m along the hill? How much work is done by the rope on the sled in this distance? What is the work done by gravity on the sled? What is the total work done?

Solution

W_f = mu_s * m g * cos phi * (d cos theta)

= 0.100 * 95 * 9.8 * cos 52 * 30 * cos 180

= -1720 = -1.7 * 10³ J

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W_r = T d cos theta

T + F_f = m g sin phi

W_r = mg ( sin phi - mu_s cos phi ) * d cos theta

= (95 * 9.8) [ sin 52 - 0.100 * cos 52 ] * 30 * cos 180

= -20288 = -2.02 * 10⁴ J

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W_g = m g * sin phi * (d * cos theta)

= 95 * 9.8 * sin 52 * 30 * cos 0

= 220008 J

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W = W_f + W_g + W_r

= 220008 + (-2.02 * 10⁴ J) + (-1.7 * 10³ )

= 0 J