(a) A 2 kg object is at the base of a 35° incline, initially at rest,...

(a) A 2 kg object is at the base of a 35° incline, initially at rest, and sits on top of a spring (k = 240 N/m) which is parallel to the inline. The spring is compressed 0.40 m from its equilibrium position and then released, propelling the object up the incline. While the object slides it experiences a frictional force of 2.00 N. Diagram this situation below. Remember to include coordinate axes.

(b) Apply conservation of energy to determine the speed of the object at the moment the spring reaches equilibrium. Explain all terms. If any terms equal zero explain why.

(c) Apply conservation of energy to determine how far up the incline the object will reach (above the spring’s equilibrium position). Explain all terms. If any terms equal zero explain why.

(d) Apply conservation of energy to determine how much the spring will compress in stopping the object as it slides back down the incline. Explain all terms. If any terms equal zero explain why.

Expert Solution

In part C max height that mass M reach from equilibrium is 1.05m and from compressed state it is 1.45m.

Mainly you have to conserve energy and focus on energy loss due to friction. And final compression is approx (0.325m - 0.335m). I got nearly 0.325 ,may be your value get slightly differ from mine but it should in range that I have provided.

genius_generous answered 5 months ago

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