Answer:
7. Net constant force down the ramp
Explanation:
After the car is released and starts moving up the ramp, the only force that is applied on the car is weight because of the gravity, we were told that the friction force is neglected. the force because of the weight is given by:
where θ is the angle of the ramp.
as you can see those values won't change, so the force remains constant down the ramp.
Answer:
Efriction = 768.23 [kJ]
Explanation:
In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.
We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.
The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.
where:
m = mass = 56 [kg]
h = elevation = 1400 [m]
v = velocity = 5.6 [m/s]
Answer:
-2000 N
Explanation:
To solve the problem, we can use the impulse theorem, which states that the impulse is equal to the change in momentum of the car:
where
F is the average breaking force
is the stopping time
m = 1000 kg is the mass of the car
is the change in velocity of the car
Solving the equation for F,