Question

A 8.75-kg block is sent up a ramp inclined at an angle θ = 30.5° from the horizontal. It is given an initial velocity v0 = 15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is μk = 0.353 and the coefficient of static friction is μs = 0.639. How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

Answer 1

14.1 meters.

Step-by-step explanation

• Weight on the block:
  `W = m \cdot g = 8.75 * 9.81 = 85.8375\;\text{N}`.
• Consider the weight on the block as two components: along the slope and normal to the slope. Magnitude of the component along the slope:
  `W_\text{along the slope} = W \cdot sin(\theta) = 43.5658\;\text{N}`.
• Normal force on the block, which is the same as the component of weight normal to the slope:
  `N = W_\text{normal to the slope} = W \cdot cos(\theta) = 73.9601\;\text{N}`.
• Kinetic friction:
  `F_\text{kinetic friction} = \mu_k\cdot N = 26.1079\;\text{N}`.
• Maximum Static friction:
  `F_\text{static friction, max} = \mu_s\cdot N = 47.2605\;\text{N}`.

`F_\text{static friction, max} > W_\text{along the slope}`. As a result, the block won't slip downwards after it comes to a stop.

As the block moves upward:

• Net force on the block
  `\sum F = F_\text{kinetic friction} +W_\text{along the slope} = -69.6737\;\text{N}`.
• Acceleration of the block:
  `{a} = \frac{{F}}{m} = -7.96271\;\text{m}\cdot\text{s}^(-2)`.

• Initial velocity
  `v_0 = 15.0\;\text{m}\cdot\text{s}^(-1)`.
• Final velocity of the block
  `v_1 = 0` as it comes to a stop, given that it doesn't slip downwards.

How far does the block move in this process?

`x = \frac{{v_1}^(2)-{v_0}^(2)}{2a} = 14.1\;\text{m}` (3 sig. fig.).