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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?

Respuesta :

Answer:

The acceleration up the ramp is 6.68 m/s².

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

[tex]F = mg[/tex]

[tex]m = \dfrac{F}{g}[/tex]

Put the value into the formula

[tex]m=\dfrac{189}{9.8}[/tex]

[tex]m=19.28\ kg[/tex]

We need to calculate the acceleration up the ramp

Using balance equation

[tex]189\cos\theta-mg\sin\theta=ma[/tex]

Put the value into the formula

[tex]189\cos28-19.28\times9.8\sin11.6=19.28\times a[/tex]

[tex]a=\dfrac{189\cos28-19.28\times9.8\sin11.6}{19.28}[/tex]

[tex]a=6.68\ m/s^2[/tex]

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

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