Answer:
568.16J
Explanation:
V = 1.4 m/s
h = 2.15m
H = 12.4m
M = 57.0kg
Fr = 41.0N
Vb = 6.80m/s
g = 9.8m/s²
K.E at the top + P.E at the top + Work done = K.E at the bottom + frictional force.
½mv² + mgh + W = ½mVb + Fr.H
(½* 57 * 1.4²) + (57 * 9.8 * 2.15) + W = (½ * 57 * 6.8²) + (41 * 12.4)
55.86 + 1202.22 + w = 1317.84 + 508.4
1258.08 + w = 1828.24
W = 1826.24 - 1258.08
W = 568.16J
The work done by the girl as the bike travels down the incline is 568.16 J.
Since A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass is 57.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.80 m/s.
Now
We know that
K.E at the top + P.E at the top + Work done = K.E at the bottom + frictional force.
So,
½mv² + mgh + W = ½mVb + Fr.H
(½* 57 * 1.4²) + (57 * 9.8 * 2.15) + W = (½ * 57 * 6.8²) + (41 * 12.4)
55.86 + 1202.22 + w = 1317.84 + 508.4
1258.08 + w = 1828.24
W = 1826.24 - 1258.08
W = 568.16J
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