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

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s