Question

1. A truck was pulling a R.V. Away from the end of the cliff. The chain broke and the RV fell to the ground. If the truck had an initial velocity of 2 m/s , what was the distance it fell from the edge of the cliff? The cliff is 280 ft tall. What would be the Force of impact? (R.V=1800 kg)

Answer 1

The force of impact, F = 17682.18 N

Step-by-step explanation:

Given,

The initial velocity of the truck and associated velocity of RV, u = 2 m/s

The height of the cliff, h = 280 ft

= 85.344 m

The mass of the RV, m = 1800 Kg

The total energy of the RV at height, 'h' with velocity, 'v', E = P.E + K.E

P.E = mgh J

= 1800 x 9.8 x 85.344

= 1505468.16 J

K.E = mv²/2

= 1800 x 2² / 2

=3600 J

Therefore total energy, E = 1509068.16 J

This is equal to the kinetic energy of RV at the impact.

The force of impact,

F = E/h

= 1509068.16 J / 85.344 m

= 17682.18 N

Hence, the force of impact of RV is, F = 17682.18 N