2016SpringMay EminationPHY 210 715A 14615qHCC. A- Question 1: (20pt) Examination

Question

2016SpringMay EminationPHY 210 715A 14615qHCC. A- Question 1: (20pt) Examination 2016Spring May PHY 2107 As shown in figure on the right, there is a spring fixed on the wall and relaxed at point Prelax. Ther 2.00 kg block sliding on the frictionless ground from Po at beginning. Then the block is released a to. The block slide freely up to an inclined plane and stops at P2, 0.30 m high above the ground. a) What is the power of the work done by the block's weight from ti to t2? (5 pt) b) What is the work done by the spring from to to t? (5 pt) c) What is the spring force constant? (5 pt) d) Which U-t graph is reasonable? (5 pt) (Define the ground level as origin. Hint: only one answer m 200kg to:0; t1 = 1.20 s,t2 = 5.00 s Pretax-Po = 2.94 m; P2-Prelax .: 1.00 m Circle One: A

Explanation / Answer

Question 2

Mbullet=m=0.1 kg

Mwood=M=10 kg

h=0.8m

Applying the law of momentum conservation for the horizontal direction:

pbefore=pafter ---> mv=(m+M)V; where V is the veolicty of the bullet-woodblock system (this is the answer for part c)

therefore, V=(m/(m+M)v (this will be the answer for part b)

During the whole process, the mechanic energy is conserved. Thus, (K+U)o=(K+U)f, (this is the answer for part a)

0.5(m+M)V^2+0=0+(m+M)gh ----> v=(m+M/m)sqrt(2gh)=(0.1kg+10kg/0.1kg)sqrt(2*9.8*0.8)= 399m/s (numerical answer for part d)

V=(0.1kg/(10.1kg)399m/s=3.95m/s (numerical answer for part b)

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