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. Determine the work done (in J) by the girl as the bike travels down the incline.

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

cant without the picture! ?

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft