Question

) A crate (mcrate = 12 kg) is sliding along a horizontal surface with an initial velocity when it goes up a ramp, the crate goes up the ramp a distance of d = 15 m where is stops a vertical height of h = 2 meters above the ground, a frictional force of 144.3 N acted on the crate over the 15 m, what was the initial velocity of the crate?

Answer 1

Here we have given that ,

mass of crate = 12kg

ramp a distance of d = 15 m

vertical height of h = 2 m

angle of inclination = sin-(2/15) =7.66 degree

to find =  initial velocity of the crate = v

frictional force = 144.3 N

so k = 144.3/12*9.81 = 1.2232

so that, on balancing the forces ,

∑F=ma

a=∑F/m

The net horizontal force ∑F is

∑F=mgsinθ +fk = mgsinθ + μkmgcosθ

a=∑F/m = gsinθ + μkgcosθ

so that , a= 13.2001 m/s2

directed down the incline, so this can also be written as

a=−13.2001 m/s2

Now, we can use the equation

(vx)^2=(v0x)^2+2ax(Δx)

to find the distance it travels up the ramp before it comes to a stop.

Here,

• Vx=0 (instantaneously at rest at maximum height)

• V0x=trying to find

• ax=−13.2001

• Δx= 15 m

Plugging in known values, we have

Vo= ( 2 * 13.2001* 15)^0.5 = 19.8998 m/s

hence the  initial velocity of the crate

is 19.8998 m/s