Question

1. A block with mass 20 kg is

sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?

Answer 1

Answer: 0.435 m

Step-by-step explanation:

Given

mass m=20 kg

initial speed u=2 m/s

coefficient of kinetic friction
`\mu_k=0.3`

deceleration which opposes the motion is given by

`\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)`

`\Rightarrow a=9.8(\sin 10^(\circ)+0.3* \cos 10^(\circ))\\\Rightarrow a=4.59\ m/s^2`

using
`v^2-u^2=2as`

`\Rightarrow s=(2^2)/(2* 4.59)=0.435\ m`