If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of
1 answer:
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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