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2 answers:
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Refer to the diagram shown below.
The weight of the block is
W = (2.41 kg)*(9.8 m/s²) = 23.618 N
The kinetic coefficient of friction is
μ = 0.521
The normal reaction is
N = f cos(59.9°) = 0.5015f N
The frictional resistive force is
R = μN = 0.521*(0.5015f N) = 0.2613f N
For dynamic force balance,
f sin(59.9°) = W + R
0.8652f = 23.618 + 0.2613f
0.6039f = 23.618
f = 39.109 N
The block moves by 1.42 m.
The work done is
W = (f sin(59.9° N)*(1.42 m)
= 39.109*sin(59.9°)*1.42
= 48.05 J
Answer: 48.0 J (nearest tenth
The weight of the block is
W = (2.41 kg)*(9.8 m/s²) = 23.618 N
The kinetic coefficient of friction is
μ = 0.521
The normal reaction is
N = f cos(59.9°) = 0.5015f N
The frictional resistive force is
R = μN = 0.521*(0.5015f N) = 0.2613f N
For dynamic force balance,
f sin(59.9°) = W + R
0.8652f = 23.618 + 0.2613f
0.6039f = 23.618
f = 39.109 N
The block moves by 1.42 m.
The work done is
W = (f sin(59.9° N)*(1.42 m)
= 39.109*sin(59.9°)*1.42
= 48.05 J
Answer: 48.0 J (nearest tenth