Question

A box that has a mass of 80 kg slides down a ramp with a 30 degree angle. The free-body diagram shows the forces acting on the box. A block is shown on a 30-degree incline with a free body diagram drawn on the block with 2 force vectors. The first vector is pointing away from the surface of the incline and perpendicular to the surface, labeled F Subscript N Baseline. The second vector is pointing straight down to the center of the earth, labeled F Subscript g Baseline. Ignoring friction and air resistance, what is the acceleration of the box, to the nearest tenth? 0.5 m/s2 4.9 m/s2 8.5 m/s2 9.8 m/s2

HOW TO DO IT

Answer 1

b

Step-by-step explanation:

b

Answer 2

B. 4.9 m/
`s^(2)`

Step-by-step explanation:

The parallel vector,
`F_(g)`, make the box to accelerate down the plane due to the presence of an unbalanced force.

The perpendicular component of the force will balance the normal force because the box can not accelerate perpendicularly to the plane.

i.e
`F_(N)` =
`F_(g)`

Since the box slides down the ramp, neglecting friction and air resistance;

`F_(g)` = ma

mg Sinθ = ma

But; m = 80 kg, θ =
`30^(0)`, g = 9.8 m/
`s^(2)`

80 × 9.8 × Sin
`30^(0)` = 80 × a

784 × 0.5 = 80 a

392 = 80 a

a =
`(392)/(80)`

= 4.9

Thus, the box accelerates down the ramp at 4.9 m/
`s^(2)`.