Question

Now the student moves the box up a ramp (with the same coefficient of friction) inclined at 11.6 ◦ with the horizontal. b) If the box starts from rest at the bottom of the ramp and is pulled at an angle of 28◦ with respect to the incline and with the same 189 N force, what is the acceleration up the ramp?

Answer 1

The acceleration up the ramp is 6.68 m/s².

Step-by-step explanation:

Given that,

Inclined = 11.6°

Pulled angle = 28°

Force = 189 N

We need to calculate the mass of the box

Using formula of force

`F = mg`

`m = (F)/(g)`

Put the value into the formula

`m=(189)/(9.8)`

`m=19.28\ kg`

We need to calculate the acceleration up the ramp

Using balance equation

`189\cos\theta-mg\sin\theta=ma`

Put the value into the formula

`189\cos28-19.28*9.8\sin11.6=19.28* a`

`a=(189\cos28-19.28*9.8\sin11.6)/(19.28)`

`a=6.68\ m/s^2`

Hence, The acceleration up the ramp is 6.68 m/s².