A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity?0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is? k=0.460 and the coefficient of static friction is? s=0.817. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

Question

Physics

Laci Dunlap

11 November, 13:31

A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity?0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is? k=0.460 and the coefficient of static friction is? s=0.817. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

+5

Answers (1)

Know the Answer?

James Wiggins · Answer 1

d = 11.79 m

Explanation:

Known data

m=4.25 kg : mass of the block

θ = 37.5° : angle θ of the ramp with respect to the horizontal direction

μk = 0.460 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W = m*g

W = 4.25 kg * 9.8 m/s² = 41.65 N

x-y weight components

Wx = Wsin θ = 41.65*sin 37.5° = 25.35 N

Wy = Wcos θ = 41.65*cos 37.5° = 33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk * N = 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax, ax = a : acceleration of the block

-Wx-f = m*a

-25.35-15.2 = (4.25) * a

-40.55 = (4.25) * a

a = (-40.55) / (4.25)

a = - 9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block:

vf²=v₀²+2*a*d Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

dа ta:

v₀ = 15 m/s

vf = 0

a = - 9.54 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15) ²+2 * (-9.54) * d

2 * (9.54) * d = (15) ²

(19.08) * d = 225

d = 225 / (19.08)

d = 11.79 m