Question

7: A 60-kg box is placed against a uniform spring with a spring constant (k) of 3300 N/m The box is pushed to the right so it compresses the spring 25-cm. When the box is released, it is propelled to the left. (a) What is the velocity of the box as it slides along the flat surface to the left after it loses contact with the spring? 350 (b) How far up the ramp will the box slide while coming to a stop? Assume the flat section has no friction and the ramp has a coefficient of kinetic friction of .18 and static friction of .23 (c) After the box stops, will it remain stopped or will it slide back down the ramp (give rationale for your answer)? 20-pts

Answer 1

From conservation energy 1/2 mv^2 = 1/2 kx^2 60 times v^2 = 3300 times 0. 25^2 v = 1. 854 m/s. (b) from conservation of energy 1/2 kx^2 = mg h + mu mg d 1/2 times 3300 times 0. 25^2 = 60 times 9. 8 times d times sin 35 degree + 0. 18 times 60 times 9. 8 times d d = 0. 233m (c) it will not stopped it side have down the sump