This HW consists of two similar problems This HW consists of two similar problem

Question

This HW consists of two similar problems

This HW consists of two similar problems-one with friction and one without. Try to solve the one without friction and understand it first, then see the effects of adding friction to energy problems. Assume a block of mass m = 1.0 kg slides down a ramp and onto a horizontal surface. The ramp has length of 1.0 meter. The block slides horizontally a further distance of 1.0 meter before striking a spring of spring constant 100.0 N/m. The ramp's incline makes an angle of 30 degrees with respect to the horizontal. Neglect friction and air resistance. Describe the transformation of energy in this system between different forms as the block slides down, compresses the spring and then rebounds. Include the cases 1) when the block is at the top of the ramp, 2) when it is sliding down the ramp, 3) when it is sliding horizontally, 4) when it compresses the spring, 5) when it rebounds back up the ramp. What is the speed of the block at the bottom of the incline? How far does the block compress the spring? The box rebounds up the ramp. How close does it come to its initial position? (Please express the result as distance along the ramp's surface.)

Explanation / Answer

a)

1) at the top of the surface the block has only gravitational potentai energy.

U = m*g*h = m*g*L/sin(30)

2) as the blcok slides its gravitational potential

energy decreases and kinetic energy increases.

at the botoom, 0.5*m*v^2 = m*g*L/sin(30)

3) when it is sliding horizonatlly it has constant kinetic energy.

KE = 0.5*m*v^2

4) when it compresses the spring kinetic energy decreases and elastic potential energy of the spring increases.

o.5*k*x^2 = 0.5*m*v^2

5) when rebounds back elastic potential energy of the spring decreases and kinetic enrgy of the spring increases.

0.5*m*v^2 = 0.5*k*x^2

b) m*g*L/sin(30) = 0.5*m*v^2

==> v = sqrt(g*h) = sqrt(9.8*1) = 3.13 m/s

c) 0.5*k*x^2 = 0.5*m*v^2

x = sqrt(m/k)*v = sqrt(1/100)*3.13 = 0.313 m = 31.3 cm

d) as the two surface are frictionless the block again comes to its initial position.

b)

1) at the top of the surface the block has only gravitational potentai energy.

U = m*g*h = m*g*L/sin(30)

2) as the blcok slides its gravitational potential

energy decreases and kinetic energy increases.

here frictional force does negative work. so

final kinetic energy is not equal to initial potential energy.


3) when it is sliding horizonatlly frictional force does negative work on the block. so, as it moves its kinetic energy decreases.


4) when it compresses the block kinetic energy decreases and elastic potential energy of the spring increases.

here also frictional force does negative work. so, the potentail energy of the spring is not same as kinetic energy of the block.

5) when rebounds back elastic potential energy of the spring decreases and kinetic enrgy of the spring increases.

here also frictional force does negative work. so, the kinetic energy of the block is not same as potentail energy of the spring.



b) a = g*sin(30) - mue*g*cos(30) = 4.05 m/s^2

v^2-u^2 = 2*a*s

v = sqrt(2*a*s) = sqrt(2*4.05*1) = 2.846 m/s

c)


0.5*m*v^2 - mue*m*g*1 - mue*m*g*x = 0.5*k*x^2


0.5*1*2.846^2 - 0.1*1*9.8 - 0.1*1*9.8*x = 0.5*100*x^2


==> 50*x^2 + 0.98*x - 3.07 = 0

soving above equation

we get, x = 0.238 m = 23.8 cm


d)0.5*k*x^2 = mue*m*g*(1+x)+[mue*m*g*cos(30)+mue*m*g*sin(30)]*L

L = 0.5*k*x^2- mue*m*g*(1+x) ]/[mue*m*g*cos(30)+ m*g*sin(30)]}

L = 0.281 m


it comes 100-28.1 = 71.9 cm close

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