A jack-in-the-box with a box of mass M and clown of mass m = 200.0 g rests on a

Question

A jack-in-the-box with a box of mass M and clown of mass m = 200.0 g rests on a horizontal table (Fig. P8.50). The clown is attached to the box by means of a lightweight, vertical spring with force constant k = 1.00 times 102 N/m. Zara pushes down on the clown with an additional force of 4 mg, further compressing the spring. When Zara removes her hand, the clown jumps upward, and the spring lifts the box off the table. What is the maximum value for the mass M of the box for this action to occur?

Explanation / Answer

Consider the situation for maximum value of M –

In this situation, total force in the vertically downward direction –

Fd = (M+m)*g

Now, we determine the maximum compression of the spring, x –

F = k*x

=> 4mg + mg = k*x

=> x = 5mg/k = (5*0.20*9.81) / (1.0x10^2) = 0.0981 m.

Again, we determine the initial velocity of mass m –

(1/2)mv^2 = (1/2)k*x^2

=> 0.2*v^2 = 10^2*0.0981^2

=> v = 2.19 m/s.

So, acceleration of mass m –

v^2 = u^2 +2*a*x

=> 2.19^2 = 2*a*0.0981

=> a = 24.52 m/s^2

Now, for lifting of mass M –

ma = M*g

=> M = ma/g = (0.2*24.52) / 9.81 = 0.50 kg

So, the maximum value of mass M = 0.50 kg.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.