What is the acceleration of the 2.0 kg block in FIGURE CP7.55 across the frictio

Question

What is the acceleration of the 2.0 kg block in FIGURE CP7.55 across the frictionless table? FIGURE CP7.56 shows a 200 g hamster sitting on an 800 g wedge- shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having mu_s = mu_k = 0 is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

Explanation / Answer

Balancing forces on 1 kg block,

1*g - T =1*a

Now acceleration of 2kg block is twice the acceleration of 1 kg block.

Balancing forces on 2 kg block,

2T = 2*2a

2g = 6a

a = g/3.

Acceleration of 2 kg block = 2a = 2g/3 = 2*9.8/3 = 6.53 m/s2

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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