A 5.0 kg block slides along a frictionless horizontal surface with a speed of 9.

Question

A 5.0 kg block slides along a frictionless horizontal surface with a speed of 9.0 m/s. After sliding a distance of 7.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 30 degrees to the horizontal. the acceleration of gravity is g=9.8 m/s2. Consider the zero of the vertical y coordinate and the zero of the potential energy at the level of the horizontal surface.

(1) What is the total mechanical energy of the block when it comes momentarily to a stop?

(2) What is the maximum height it reaches?

Explanation / Answer

Data: Mass of the block, m = 5 kg Initial speed, v = 9 m/s Distance of sliding, s = 7 m Angle of inclination of the ramp, = 30 deg Let x = Distance traveled along the ramp Solution: Vertical height, h = x sin Potential energy at the maximum height = Kinetic energy at the bottom m g h = (1/2) m v^2 2 g h = v^2 2 * 9.8 * x * sin 30 = 9^2 x = 8.26 m Ans: Distance traveled along the ramp, x = 8.26 m the maximum height it reaches is 4.13 m And when it stops it has a kinetic energy of 0, and a potential energy of 202.37 J

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