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7. [Honors] An initially stationary 15.0−kg15.0-\mathrm{kg} crate of cheese wheels is pulled, via a cable, a distance 5.70 m up a frictionless ramp, to a height hh 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 m/s5 \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

STEP 1

1. The ramp is frictionless.
2. The crate is initially stationary.
3. The crate is pulled up a distance of 5.70 m 5.70 \, \text{m} to a height h=2.50 m h = 2.50 \, \text{m} .
4. The crate has a mass of 15.0 kg 15.0 \, \text{kg} .
5. The crate travels with constant speed.
6. The gravitational acceleration g=9.8 m/s2 g = 9.8 \, \text{m/s}^2 .

STEP 2

1. Calculate the work done by the gravitational force.
2. Calculate the work done by the force from the cable.
3. Calculate the total work done on the crate if it reaches 5 m/s 5 \, \text{m/s} after 5.7 m 5.7 \, \text{m} .
4. Determine the force of tension from the total work done.

STEP 3

Calculate the work done by the gravitational force:
The work done by gravity is given by:
Wgravity=−mgh W_{\text{gravity}} = -mgh
Substitute the given values:
Wgravity=−(15.0 kg)(9.8 m/s2)(2.50 m) W_{\text{gravity}} = -(15.0 \, \text{kg})(9.8 \, \text{m/s}^2)(2.50 \, \text{m})
Wgravity=−367.5 J W_{\text{gravity}} = -367.5 \, \text{J}

STEP 4

Calculate the work done by the force from the cable:
Since the crate travels with constant speed, the work done by the cable must exactly oppose the work done by gravity:
Wcable=−Wgravity W_{\text{cable}} = -W_{\text{gravity}}
Wcable=367.5 J W_{\text{cable}} = 367.5 \, \text{J}

STEP 5

Calculate the total work done on the crate if it reaches 5 m/s 5 \, \text{m/s} after 5.7 m 5.7 \, \text{m} :
The kinetic energy at 5 m/s 5 \, \text{m/s} is:
KE=12mv2 KE = \frac{1}{2}mv^2
Substitute the given values:
KE=12(15.0 kg)(5 m/s)2 KE = \frac{1}{2}(15.0 \, \text{kg})(5 \, \text{m/s})^2
KE=187.5 J KE = 187.5 \, \text{J}
The total work done is the change in kinetic energy plus the work done against gravity:
Wtotal=KE+Wgravity W_{\text{total}} = KE + W_{\text{gravity}}
Wtotal=187.5 J+367.5 J W_{\text{total}} = 187.5 \, \text{J} + 367.5 \, \text{J}
Wtotal=555 J W_{\text{total}} = 555 \, \text{J}

STEP 6

Determine the force of tension from the total work done:
The force of tension does work over the distance of 5.70 m 5.70 \, \text{m} :
Wcable=Ftension×d W_{\text{cable}} = F_{\text{tension}} \times d
555 J=Ftension×5.70 m 555 \, \text{J} = F_{\text{tension}} \times 5.70 \, \text{m}
Solve for Ftension F_{\text{tension}} :
Ftension=555 J5.70 m F_{\text{tension}} = \frac{555 \, \text{J}}{5.70 \, \text{m}}
Ftension=97.37 N F_{\text{tension}} = 97.37 \, \text{N}
The force of tension is:
97.37 N \boxed{97.37 \, \text{N}}

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