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A 18 kg box slides from rest down a ramp inclined at 25° to the horizontal onto a spring with a spring constant 740 N/m
as shown in the diagram. The spring is compressed 0.32 m before the box stops. Determine how far along the ramp the
box will slide before it stops. Assume friction is negligible.

Respuesta :

onyebuchinnaji

The distance traveled by the box along the ramp is 0.51 m.

Distance traveled by the box

Apply the principle of conservation of energy;

Potential energy of the box at top of incline = Elastic potential energy at bottom

mgh = ¹/₂kx²

where;

  • h is the height of the incline
  • m is mass of the box
  • x is the compression of the spring
  • k is spring constant

h = L sinθ

where;

  • L is length of the incline = distance traveled by the box

mg(L sinθ) = ¹/₂kx²

(18)(9.8)(L sin25) = ¹/₂(740)(0.32)²

74.55 L = 37.89

L = 37.89 / 74.55

L = 0.51 m

Thus, the distance traveled by the box along the ramp is 0.51 m.

Learn more about distance here: https://brainly.com/question/2854969

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