Question

problem reference 6.3
a 6.00 - kg box slides from rest down a frictionless incline from a height of 5.00 m. a constant frictional force, introduced at point a, brings the block to rest at point b, 19.0 m to the right of point a.
what is the initial gravitational potential energy of the box at the top of the incline, before it is released?
75.0 j
294 j
58.8 j
30.0 j

Explanation:

Step1: Recall gravitational potential energy formula

The formula for gravitational potential energy is \(U = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity (\(g = 9.8\space m/s^{2}\)), and \(h\) is height.

Step2: Substitute values into formula

Given \(m = 6.00\space kg\), \(h=5.00\space m\), and \(g = 9.8\space m/s^{2}\). Then \(U=(6.00\space kg)\times(9.8\space m/s^{2})\times(5.00\space m)\)

$$
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Answer:

294 J