can you please help me with this? A block with mass m = 12 kg rests on a frictio

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can you please help me with this?

A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4457 N/m after being compressed a distance x1 = 0.475 m from the spring's unstretched length. The floor is frictionless except for a rough patch a distance d = 2.6 m long. For this rough path, the coefficient of friction is mu k = 0.49. How much work is done by the spring as it accelerates the block? What is the speed of the block right after it leaves the spring? How much work is done by friction as the block crosses the rough spot? What is the speed of the block after it passes the rough spot? Instead, the spring is only compressed a distance X2 = 0.112 m before being released. How far into the rough path does the block slide before coming to rest? What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released? If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is: A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4457 N/m after being compressed a distance x1 = 0.475 m from the spring's unstretched length. The floor is frictionless except for a rough patch a distance d = 2.6 m long. For this rough path, the coefficient of friction is muk = 0.49. How much work is done by the spring as it accelerates the block? What is the speed of the block right after it leaves the spring? How much work is done by friction as the block crosses the rough spot? What is the speed of the block after it passes the rough spot? Instead, the spring is only compressed a distance X2 = 0.112 m before being released. How far into the rough path does the block slide before coming to rest? What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released? If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is:

Explanation / Answer

1. Work done by spring is 0.5*k*x*x where k is spring constant, x is compression
So, result is 0.5*4457*0.475*0.475= 502.8 J

2. Work done by spring is KE of block
so 0.5*m*v*v = 502.8 J
means v*v = 502.8*2/m
implies v = sqrt (83.8) which is 9.15 m/s

3. Work done by friction = mu*m*g*d = 0.49*12*9.8*2.6 = 149.8 J

4. Speed after block passes rough patch:
   Applying conservation of energy
0.5*m*V*V= 502.8-149.8
   or V = sqrt (353*2/12) which is sqrt(58.83) or 7.67 m/s

5. spring compressed by 0.112 m
   work done = 0.5*4457*0.112*0.112= 27.95 J
mu*x*m*g = 27.95
or 0.49*12*9.8*d = 27.95
or d = 0.48 m

6. Work done by friction is 149.8 J
so, compression length required is sqrt (149.8*2/4457) = 0.73 m

7. Spring compressed 3 times farther and released versus original situation
   New situation : 0.5*k*9*x*x vs old situation 0.5*k*x*x
So factor 9 comes into picture and work done is nine times greater

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