23. A spring (k-600 N/m) is placed in a vertical position with its lower end sup

Question

23. A spring (k-600 N/m) is placed in a vertical position with its lower end supported by a horizontal surface. The upper end is depressed 20 cm, and a 4.0 kg block is placed on the depressed spring. The system is then released from rest. How far above the point of release will the block rise? a) 46 cm. b) 36 cm. c) 41 cm d) 31 cm e) 20 cm 24. A constant force of 20 N is applied to an object of mass 8.0 kg at an angle of 25 degree with the horizontal. What is the work done by this force on the object if it causes a displacement of 2.0 m along the horizontal direction? a) 40 J b) 0J c) 36J d) 17 J 25. An object with mass of 10 kg is moving along a horizontal surface. At a certain point it has a speed of 16 m/s. If the coefficient of friction between the object and the surface is 0.40, how far will the object go before it comes to a stop? Use g-10 m/s2. a) 82 m b) 61 m c) 32 m d) 23 m 26. A 4.0 kg particle is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The particle rebounds with a speed of 3.0 m/s. What is the magnitude of the impulse delivered to the particle? a) 24 Ns b) 32 Ns c) 40 Ns d) 30 Ns e) 8.0 N 27. Two ice skaters push off against one another starting from a stationary position. The 45 kg skater acquires a speed of 0.375 m/s What speed does the 60 kg skater acquire? a) 0.500 m/s b) 0.281 m/s c) 0.375 m/s d) 0m/s 28. Att-0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity of 2.0 rad/s. Two seconds later it has turned through 5.0 complete revolutions. What is the angular acceleration of this wheel? a) 17 rad/s2 b) 14 rad/s2 c) 20 rad/s2 d) 23 rad/s2 e) 13 rad/s2

Explanation / Answer

(23) Suppose ‘h’ is the height raised by the block.

use conservation of energy –

(1/2)*k*x^2 = m*g*h

=> h = (k*x^2) / (2*m*g) = (600*0.20^2) / (2*4*9.8) = 0.31 m = 31 cm

So, option (d) is the correct answer.

(24) Component of force in the direction of displacement –

Fx = F*cos25 = 20*cos25 = 18.13 N

So, work done, W = Fx * d = 18.13 * 2.0 = 36.26 J

This can be approximated to 36 J

Hence, option (c) is the correct answer.

(25) De-acceleration of the object due to friction, a = µ*g = 0.40*10 = 4.0 m/s^2

So, distance covered by the object before stopping, d = v^2 / (2a) = 16^2 / (2*4) = 32 m

Hence, option (c) is the correct answer.

(26) Magnitude of impulse = change in the momentum = mv1 – mv2

                          = 4*5 – 4*(-3) = 20 + 12 = 32 N.s

Hence option (b) is the correct answer.

(27) Apply conservation of momentum.

Initial momentum = 0

So, we have –

45*0.375 + 60*v = 0

=> v = -(45*0.375) / 60 = -0.281 m/s

Option (b) is the correct answer.

(28) Convert 5.0 revolutions in radians.

5 rev = 5*2*pi = 31.41 rad.

so we have –

31.41 = 2*2 + 0.2*a*2^2

=> a = (31.41 – 4) / 2 = 13.7 rad/s^2 = 14 rad/s^2

Hence option (b) is the correct answer.

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