Answer:
F = -4567.40 N
Explanation:
Given that,
The power developed by the engine, P = 196 hp
1 hp = 746 W
196 hp = 146157 W
Speed of the car, v = 32 m/s
Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :
F = -4567.40 N
So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²