PHYSIC RAMP QUESTION

Question

A block is at the base of a ramp of angle 35 degrees. The ramp is 5m long and has a coefficients of static and kinetic friction of 0.6 and 0.2 respectively. If the block is initially given a velocity of 12 m/s up the ramp, how far from the ramp will the block land?

Answer 1

x = 5 m, θ = 35°, v₀ = 12 m/s, μₖ = 0.2

Loss of speed going up the ramp

deceleration due to gravity = –g sin θ

deceleration due to friction = –μₖ g cos θ

total deceleration a = –g (sin θ + μₖ cos θ)

v₀ = initial speed

v₁ = speed at the top of the ramp

v₁² = v₀² + 2ax = v₀² – 2 g (sin θ + μₖ cos θ)x = (12)² – 2(9.8)[(sin(35) + (0.2)cos(35)](5)

= 144 – 98 (sin(35) + (0.2)cos(35)) = 71.734

v₁ = √(71.734) = 8.4696 m/s

Projectile Motion from the end of the ramp

Components of velocity v₁

v₁ₓ = v₁ cos θ = 8.4696 cos(35) = 6.9379 m/s

v₁ᵧ = v₁ sin θ = 8.4696 sin(35) = 4.8580 m/s

Time of flight (t)

The height of the ramp (y₀ below) is 5 sin(35) from trigonometry.

y – y₀ = v₁ᵧt – ½gt² which means that –5sin(35) = –2.8679 = 4.8580 t – 4.9 t²

So we solve the quadratic equation

4.9 t² – 4.8580 t – 2.8679 = 0

to get t = 1.4073 s

Distance traveled

d = v₁ₓ t = (6.9379(1.4073) = 9.76 m

Answer (rounded to nearest tenth)

The block will land 9.8 m from the ramp.