A 11 kg box slides down a long, frictionless incline of angle 30°. It starts fro

Question

A 11 kg box slides down a long, frictionless incline of angle 30°. It starts from rest at time t-0 at the top of the incline at a height of 23 m above ground (a) What is the original potential energy of the box relative to the ground? (b) From Newton's laws, find the distance the box travels in 1 s and its speed at t = 1 s. m/s (c) Find the potential energy and the kinetic energy of the box at t = 1 s. (kinetic energy) (potential energy) (d) Find the kinetic energy and the speed of the box just as it reaches the bottom of the incline. m/s

Explanation / Answer

a) Potential energy of the box relative to the ground = m*g*h

= 11*9.8*23

= 2479 J

b) net force acting on the box, Fnet = m*g*sin(30)

m*a = m*g*sin(30)

a = 9.8*sin(30)

= 4.9 m/s^2

distance traveled in 1s,

d = (1/2)*a*t^2

= (1/2)*4.9*1^2

= 2.45 m

seed at t = 1s,

v = vo + a*t

= 0 + 4.9*1

= 4.9 m/s

c) Potential energy = m*g*h - m*g*d*sin(30)

= 11*9.8*23 - 11*9.8*2.45*0.5

= 2347 J

KE = (1/2)*m*v^2

= (1/2)*11*4.9^2

= 132 J

d) KEf = PEi

= 2479 J

KEf = (1/2)*m*vf^2

==> vf = sqrt(2*KEf/m)

= sqrt(2*2479/11)

= 21.2 m/s

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