I really need help with this questions (you can answer in formula but it would be better to explain how to get that answer)

Question

1.) A 20-kg boy runs at 5.0 m/sec and lands on a sled whose mass is 7.0 kg. Neglecting friction, what is the speed of the sled and boy together?

2.) Calculate the kinetic energy of 1.00 kg of water after falling over a waterfall in 3.10 seconds

3.) A truck with a mass of 3000 kg accelerates to a complete stop in 10 seconds from a speed of 30 m/sec. What force used to bring the truck to a stop this quickly?

4.) What force is necessary to lift a 200 kg rock with a bar 6.00 m long when the fulcrum is placed 1.50 m from the point of contact?

Thank you!

Answer 1

1) The momentum of the 20kg boy is conserved after hopping on the sled. He increases the momentum of the sled since it is initially at rest, but they will both move at a slower velocity since their overall mass is combined.

m_boy·v_boy = (m_boy+m_sled)·v_boy&sled

20(5) = (20 + 7)·v_boy&sled

100=27·v_boy&sled

(100/27) = v_boy&sled

v_boy&sled = 3.704 m/s

2) The Kinetic Energy of the water is given by the formula:

KE = (1/2)·m·v²

We can find the velocity of the water after falling for 3.10 seconds from the freefall velocity kinematic equation:

v = g·t +v_o , where g is the acceleration due to gravity, g = 9.8 m/s²

v = 9.8(3.10) = 30.38m/s

Thus,

KE = (1/2)(1)(30.38)² = 461.47J (Joules)

3) The force would be calculated from Newton's 2nd Law:

F = m·a·

and the acceleration would be found from the velocity kinematic equation,

v = a·t + v_o , where the final velocity, v = 0 m/s since the truck comes to rest,

and the initial velocity v_o= 30 m/s

We get a = -v_o/t = -30 /10 = -3 m/s², (negative sign indicates an acceleration opposite to direction of velocity, or a deceleration) thus the Force F is:

F = 3000(-3) = -9000N (negative sign indicates a Decelerating Force acting on the truck). The magnitude of this force |F| = 9000N

4) The 200kg rock is on one end of the 6m long bar and the force is applied on the other end. The fulcrum (center of rotation or hinge) is 1.5m from the rock, meaning that the force applied is 4.5m away from the fulcrum, on the other side. In order to lift the rock, the force applied can be smaller than the weight of the 200kg rock since it is a farther distance away. A smaller force can generate the same amount of torque provided the distance to the center of rotation is farther since:

T = F×r

Thus the torque generated by the weight of the rock must be equal and opposite to the torque generated by the force. The sum of the torques around the fulcrum must be equal to zero (for static equilibrium)

∑T_fulcrum = 0 , taking counterclockwise torques as positive

F_applied·(4.5m) = m_rock·g·(1.5m)

F_applied = 200(9.8)(1.5)/4.5) = 653.3N

Thus, the minimum force necessary should be greater than 653.3N