Mechanics, help?

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1 Answer
May 8, 2018

See below

Explanation:

Let's start out by determining the force of weight, force normal and force parallel down the slope:

Fw=50kg9.8ms2=490N

Fn=50kg9.8ms2cos(10)=482.6N

Fp=50kg9.8ms2sin(10)=85.1N

So we have the force parallel that want to pull the box down the ramp, so we know that the force of static friction must be at least equal to force parallel, to make the object not move:

FnusFp

usFpFn

us85.1 N482.6 N

us0.176

Ok, so this girl is pushing down the box down the ramp, the force parallel will work in her favor, while the force of friction will not so:

FF=Fnuk

FF=482.6N0.19=91.694N

Fnet=50N+85.1N91.694N=43.406N

The work energy conservation theorem states:

W=KEfKEi

The initial KE energy is 0 J as the box is not moving so:

(43.406N)(5m)=12(50kg)v2

v=2.95ms

The third part I'm not entirely sure of but, 20 below the horizontal would still mean that the inclination of the ramp, from the bottom of the ramp at the other horizontal is 20

a=Fnetm

Fp=50kg9.8ms2sin(20)=167.59N

Fn=50kg9.8ms2cos(20)=460.45N

Ff=460.450.19=87.486N

Fnet=167.59N87.486N=80.104N

a=80.104 N50 kg

a=1.6ms2