how far up the incline does the sled move ?

Question

in a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3m/s up a 28 degree incline track. the combined mass of monkey and sled is 22 kg, and the coefficient of kinetic friction between the sled and incline is .45, the acceleration of gravity is 9.8 m/s^2. how far up the incline does the sled move ? answer in units of meters.

Answer 1

Ok, here you just need to draw the monkey and the sled at the bottom of the incline track, Now draw the weight of the monkey and the sled :

Weight = 22*9.8 = 215.6 N

Find the normal force, decomposing the weight : 215.6*cos28 = 190.36

And find the force that is parallel to the incline, and it's opposite to the movement : 215.6*sin28 = 101.2

With the normal force, we can find the force of friction :

f = 190.36*0.45 = 85.6

Let's find the desacceleration :

85.6 + 101.2 = 22a

a = 8.5 m/s^2

So, we have the initial speed : v = 3 m / s

desacceleration = 8.5 m/s^2

The sled is gonna stop when it's speed is cero, so the final speed will be : 0 m/s

using the formula :

Vf^2 = Vo^2 - 2ae

0 = 9 - 2*8.5*e

e = 0.53 meters

Hope that might help you

Answer 2

regularly occurring reaction (perpendicular to floor) is mg Cos ? = 16 x 9.80 one x Cos 22 = one hundred forty 5.5 N Friction stress = µR = 0.2 x one hundred forty 5.5 = 29.11N To push weight adverse to gravity, stress = mgSin ? = 16 x 981 x Sin 22 = fifty 8.8 N therefore finished opposing stress = 29.11 + fifty 8.8 = 87.9N Opposing stress = mass x deceleration (Newton's 2d regulation) - 87.9 = 16 a a = - 5.40 9 m/s^2 Vf = 0 Vi = 3 s = ? Vf^2 = Vi^2 - 2 x 5.40 9 x s 0 = 9 - 10.99s 10.99s = 9 s = 9 / 10.ninety 9 = 0.819 metres up the incline