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.

(A) What is the kinetic energy of the block as it moves along the horizontal surface?

(B) What distance along the ramp does the block slide before coming momentarily to a stop?

(C) What is the potential energy of the block when it comes momentarily to a stop?

Explanation / Answer

A .

K = 1/2 m v^2

K = 0.5 * 5 * 9^2 = 202.5 J <--------ans

B .

let the distance along ramp be x ,

then h/ x = sin 30 , h is the vertical distacne ,

K + P = 0 ;

m /2 [ v2^2 - v1^2 ] + mgh = 0 ;

1 /2 [ v2^2 - v1^2 ] + gh = 0 ;

1 /2 [ 0^2 - 9^2 ] + 9.8 * h = 0 ;

1 /2 [ 0^2 - 9^2 ] + 9.8 * [ x sin 30 ] = 0 ;

x= 8.2653 m <--------ans

C .

P = mgh = mg x sin 30

P = 5 * 9.8 * 8,2653 * sin 30 ;

P = 202.5 J <-----------ans

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