physics friction and acceleration

Question

a car of mass 1000kg is traveling at a speed of 45m/s at the top of a 25 degrees incline. The surface of the incline has uk= 0.3 Find work done by gravity as a car moves tot the bottom of the incline. work done by friction. Speed of the object.

Answer 1

Given: m = 1000kg, v₀ = 45 m/s, θ = 25º, u_k = 0.3

work = (force) * (distance)

work done by gravity = F_gravity * d = (m * g * sinθ) * d = (1000 kg)(9.8 m/s²)(sin 25º) * d

work done by friction = F_friction * d = (u_k * F_normal) * d = (u_k * m * g * cosθ) * d = (0.3)(1000 kg)(9.8 m/s²)(cos 25º) * d

To determine the velocity at the bottom, use the conservation of total energy. At the top, you have kinetic energy (because it is moving) and potential energy (due to gravity). At the bottom, you just have kinetic energy (because it is still moving) and all of the potential energy is converted into kinetic energy. By setting the final kinetic energy equal to the initial kinetic and initial potential energy, you get the following equation:

(.5mv²)_final = (.5mv²)_initial + (mgh)_initial

v_final = √[(.5mv₀² + mgh₀) / .5m] = √(v₀² + 2gh₀) = √[ (45 m/s)² + 2(9.8 m/s²)(h₀) ]

To solve the work equations, you need to know the actual distance the car travels. To solve for the final velocity, you need to know the change in vertical height.