Please see below help would be appreciated, thanks in advance! What acceleration

Question

Please see below help would be appreciated, thanks in advance!

What acceleration is required to accelerate a 2 kg box from rest to 5 m/s over a distance of 2 meters on a frictionless surface? What force (in the direction shown) is required to accelerate a 2 kg box from rest to 5 m/s over a distance of 2 meters on a frictionless surface? What force (in the direction shown) is required to accelerate a 2 kg box from rest to 5 m/s over a distance of 2 meters on a rough surface with a coefficient of kinetic friction of 0.5 between the surface and the box? What acceleration is required to accelerate a box from rest to 5 m/s over a distance of 2 meters on a frictionless surface up a ramp of 30 degree ? What force (in the direction shown) is required to accelerate a box from rest to 5 m/s over a distance of 2 meters on a frictionless surface up a ramp of 30 degree ? What force (in the direction shown) is required to accelerate a box from rest to 5 m/s over a distance of 2 meters on a rough surface up a ramp of 30 degree with a coefficient of

Explanation / Answer

a) v^2 = v0^2 + 2 a x

5^2 = 0^2 + 2*a*2

25 = 4 a

a = 6.25 m/s^2

b) F = ma = 2*6.25 = 12.5 N

c)

now F - friction = m a

friction = u m g

F = um g + ma = 0.5*2*9.81+12.5=22.31 N

d)same as part a

a = 6.25 m/s^2

e) now F - mg sin theta = m a

F = mg sin theta + m a = 2*9.81*sin(30 degrees) +12.5= 22.31 N

f) so sum forces perp to slope

N - m g cos theta = 0

N = mg cos theta
so friction = u N = u m g cos theta

F - m g sin theta - friction = ma

F = 2*9.81*sin(30 degrees) +12.5 + 0.5*2*9.81*cos(30 degrees)= 30.8 n

g)
now when you sum perp to slope

N - mg cos theta - F sin theta = 0

N = m g cos theta + F sin theta

so now sum with slope

F cos theta - m g sin theta - u ( mg cos theta + F sin theta) = ma

F*cos(30 degrees) - 2*9.81*sin(30 degrees) - 0.5*(2*9.81*cos(30 degrees) +F*sin(30 degrees) ) = 2*6.25

F=50 N

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