A 500 N Crate kept on the top of a 15° sloping surface is pushed down the plane with an initial velocity of 20m/s. If μs = 0.5 and μk = 0.4, determine the distance travelled by the block and the time it will take as it comes to rest.

Given: Weight of crate = 500 N

Initial velocity(u) = 20 m/s

μs = 0.5

μk = 0.4

θ = 15°

Final velocity (v) = 0 m/s

To find: Distance travelled by the block Time it will take before coming to rest

#### Solution

Mass `(M) W/g`

`500/9.81`

`=50.9684 kg`

Normal reaction (N) on the crate = 500 cos 15

Kinetic friction (Fk) = μk x N

=` 0.4xx500 cos 15`

= `193.1852 N`

Let T be the force down the incline Taking forces towards right of the crate as positive and forces towards left as negative

`T+Fk=500 sin 15`

∴ `T=500 sin 15-193.1852`

∴ `T=-63.7756 N`

By Newton’s second law of motion

`a=F/m`

∴ `a= -63.7756/50.96684=-1.2513 m/s^2`

Using kinematical equation:

`v^2=u^+2as`

∴` 0=202-2xx1.2513 xx s`

∴ `s=159.8366 m`

Using kinematical equation:

V=u+at

∴ `0=20-1.2513t`

∴ `t=15.9837 s`

∴ Distance travelled by the block before stopping = 159.8366 m ∴ Time taken by the block before stopping = 15.9847 s |