Assignments
1 answer:
Hello there!
I hope you and your family are staying safe and healthy during this unprecendented time.
A) What is the work done?
Answer: We need to use the formula
B) What is the work done on the cart by the gravitational force?
Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:
C) What is the work done on the cart by the shopper?
This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force
D) Find the force the shopper exerts, using energy considerations.
E) What is the total work done?
You just need to add them:
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Refer to the free body diagram shown melow.
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N