Mass and Spring Problem, Please Help!

Question

A 1.50 kg mass slides to the right on a surface having a coefficient of kinetic friction µk = 0.250 (Fig. P8.62). The mass has a speed of vi = 3.30 m/s when contact is made with a light spring that has a spring constant k = 50.0 N/m. The mass comes to rest after the spring has been compressed a distance d. The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched position. Finally the mass comes to rest a distance D to the left of the unstretched spring.

http://www.webassign.net/sb5/p8-62.gif

(a) Find the distance of compression d.

(b) Find the speed v of the mass at the unstretched position when the system is moving to the left.

(c) Find the distance D between the unstretched spring and the point at which the mass comes to rest.

Please help, this is really bugging me, if you could help walk me through the problems, it would be greatly appreciated. Although if it's too much, the question of the most importance is #3, so I just really need a walkthrough of that last question, though the others are also appreciated.

Answer 1

a) Let N1 = upward normal force on the block by the surface.

There is no acceleration in vertical direction. Therefore, total force in vertical direction = 0

Or, N1 - mg = 0, where m is the mass of the block.

Or, N1 = mg

Or, N1 = 1.50 * 9.8 N = 14.7 N

Let f = force of friction on the block by the surface.

f = µk * N1 = 0.250 * 14.7 N

f = 3.675 N---------(1)

Consider gravitational potential energy on the surface as zero.

Before collision between block and spring,

K.E. of block = 1/2 * m * vi^2 = 1/2 * 1.5 * 3.30^2 = 8.17 Joule---(2)

K.E. of spring = 0

P.E. of block = 0

P.E. of spring = 0

M.E. of block + spring system = 8.17 J-----------(3)

After the spring is compressed to distance d and the block comes to rest

K.E. of block = 0

K.E. of spring = 0

P.E. of block = 0

P.E. of spring = 1/2 * k * d^2 = 1/2 * 50 * d^2 = 25d^2

M.E. of system = 25d^2-------------(4)

From (3) and (4),

change in M.E. = 25d^2 - 8.17

This is equal to work done by friction. Work done by friction equals -f*d (minus sign because friction is opposite to displacement)

= - 3.675d (using (1))

Therefore, 25d^2 - 8.17 = - 3.675d

Or, 25d^2 + 3.675 d - 8.17 = 0

This is a quadratic equation of form ax^2 + bx + c, where x = d, a = 25, b = 3.675, c = -8.17

Discriminant D = b^2 - 4ac = 3.675^2 - 4 * 25 * (-8.17) = 830.5

d = (-b +- sqrt(D))/(2a) = (3.675 +- sqrt(830.5))/(2*25)

= (3.675 +- 28.8)/50

d cannot be negative.

Therefore, d = (3.675 + 28.8)/50 = 0.65 m-----------(5)

Ans: 0.65 m

b) Consider two states. One when the block is moving to the right and the spring is unstretched. Second when the block is moving to the left and the spring is unstretched. P.E. remains same.

K.E. changes by 1/2 m * v^2 - 1/2 m * vi^2

So, M.E. changes by 1/2 m * v^2 - 1/2 m * vi^2

Work done by friction = -2fd = -2*3.675*0.65 = -4.78 J [using (5)]

Therefore, 1/2 m * v^2 - 1/2 m * vi^2 = -4.78

Or, 1/2 m v^2 = 1/2 m * vi^2 - 4.78

= 8.17 - 4.78 [using (2)]

1/2 m v^2 = 3.39 J

Or, 1/2 * 1.5 * v^2 = 3.39-----------------(6)

Or, 0.75 * v^2 = 3.39

Or, v^2 = 3.39/0.75 = 4.52

Or, v = sqrt(4.52) = 2.126 m/s

Ans: 2.126 m/s

c) P.E. remains same i.e. zero

Initial K.E. = 3.39 J        [from (6)]

Final K.E. = 0

Therefore, change in M.E. = 0 -3.39 J = -3.39 J

Work done by friction = -fD

Therefore, -fD = 3.39

Or, D = 3.39/f = 3.39/3.675 = 0.922 m

Ans: 0.922 m

Please check the calculations.