Math / Calculus

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7. [Honors] An initially stationary $15.0-\mathrm{kg}$ crate of cheese wheels is pulled, via a cable, a distance 5.70 m up a frictionless ramp, to a height $h$ of 2.50 m . The crate travels with constant speed. (A) How much work is done on the crate by the gravitational force during the lift? (B) How much work is done on the crate by the force from the cable during the lift? (C) If the crate started from rest and is moving $5 \mathrm{~m} / \mathrm{s}$ after 5.7 m , what is the total work done on the crate? (D) From your calculations from C, what is the Force of Tension?

Studdy Solution
Determine the force of tension from the total work done:
The force of tension does work over the distance of $5.70 \, \text{m}$ :
$W_{\text{cable}} = F_{\text{tension}} \times d$
$555 \, \text{J} = F_{\text{tension}} \times 5.70 \, \text{m}$
Solve for $F_{\text{tension}}$ :
$F_{\text{tension}} = \frac{555 \, \text{J}}{5.70 \, \text{m}}$
$F_{\text{tension}} = 97.37 \, \text{N}$
The force of tension is:
$\boxed{97.37 \, \text{N}}$

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