A box is sliding down a ramp sloping 25° above the horizontal. The coefficient of kinetic friction is 0.15. What is the acceleration?
The answer is 2.8 m/s^2.
How can this be solved with out mass?
I will be happy to critique your work or thinking. Break the weight of the box, mg, into normal components, and a component goingdown the slide.
mass will divide out of the equation, as it is in all terms.
To find the acceleration of the box sliding down the ramp without knowing its mass, we can use the concepts of forces and Newton's second law.
1. First, draw a free body diagram of the box on the slope. The forces acting on the box are its weight (mg), the normal force (N) exerted by the ramp, and the force of kinetic friction (fk).
2. Decompose the weight force into its components. The weight force can be split into two perpendicular components: the component perpendicular to the ramp (N), which balances the normal force, and the component parallel to the ramp (mg*sinθ), which contributes to the acceleration of the box down the slope.
3. The force of kinetic friction can be calculated as fk = μk*N, where μk represents the coefficient of kinetic friction.
4. Apply Newton's second law in the direction parallel to the slope: ΣF = ma. The net force acting on the box parallel to the slope is the force parallel to the slope (mg*sinθ) minus the force of kinetic friction (fk).
mg*sinθ - fk = ma
5. Substitute the expression for fk from step 3 into the equation:
mg*sinθ - μk*N = ma
6. The normal force (N) can be obtained by balancing the forces in the vertical direction. On an inclined plane, the normal force can be calculated as N = mg*cosθ.
7. Substitute the expression for N into the equation:
mg*sinθ - μk*(mg*cosθ) = ma
8. Simplify the equation by canceling out the mass (m) term:
g*sinθ - μk*g*cosθ = a
9. Finally, plug in the known values: θ = 25°, μk = 0.15, and g = 9.8 m/s^2, and solve for the acceleration (a):
a = g*(sinθ - μk*cosθ)
a = 9.8*(sin(25°) - 0.15*cos(25°))
a ≈ 2.8 m/s^2
Therefore, the acceleration of the box sliding down the ramp is approximately 2.8 m/s^2.