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UNIVERSITI TUNKU ABDUL RAHMAN

Centre : Centre for Foundation Studies (CFS) Unit Code : FHSP 1014Course : Foundation in Science Unit Title : Physics IYear/ Trimester Session

:

:

Year 1 / Trimester 1

2014/05

Lecturer : Ms.Nurfadzilah Mr Chin Kong Yew

Additional Questions 5: Energy- W ork Theorem & Power

Q1. (a) A 2.00 kg block is released from rest at the top of an incline plane as shown in Figure 1. The coefficient of kinetic friction is 0.200.

(i) What is the work done by the kinetic friction? (ii) What is the change in potential energy when the block reaches the

bottom of the incline? (iii) Determine the final velocity of the block at the bottom of the incline.

[Answer: - 81.5 J, 235.2 J, 12.4 m/s]

Q2.

A ballistic pendulum is illustrated in Figure 1. A 40.0 g ball is caught by a 500 g suspended mass. After impact, the two masses rise a vertical distance of 45.0 mm. Find the velocity of the combined masses just after impact?

[Answer: 0.939 m s–1]

Q3. An 85 N box of oranges is being pushed across by a horizontal force. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downwards. Calculate the coefficient of kinetic friction between the box and the floor. [Answer: 0.253]

1

Figure 1

30.0o

12.0 m

Figure 1

Q4. The figure shows a spring of negligible mass that is compressed 10 cm. A 100 gram mass is placed at the end of the spring. The spring constant k = 500 N/m. When the spring is released, the 100 g mass is propelled by the spring along the horizontal rough surface and moves for a distance of 8.00 m before coming to a stop. Calculate the initial velocity of propulsion. What is the coefficient of friction of the surface?

[Answer: 7.07 m/s, 0.319]

Q5.

A 60 kg trampoline artist jumps vertically upwards from the top of a platform with a speed of 6 ms–1 as shown in Diagram 2.2.

(i) Determine his velocity as he lands on the trampoline.

(ii) If the trampoline behaves like a spring with a force constant of 6.8 10 4 Nm–1, how far will he depress it?

[Answer: 9.74 m/s, 0.298 m]

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Diagram 2.2

UNIVERSITI TUNKU ABDUL RAHMAN

Centre : Centre for Foundation Studies (CFS) Unit Code : FHSP 1014Course : Foundation in Science Unit Title : Physics IYear/ Trimester Session

:

:

Year 1 / Trimester 1

2014/05

Lecturer : Ms.Nurfadzilah Mr Chin Kong Yew

Answer Guidelines f or Additional Questions 5

1. Work done by friction = (Fcos)x = - (0.200)(2 x 9.8 cos 30.0o N)(24.0 m) = - 81.5 JPE = mgh = (2.0)(9.80)(12.0) = 235.2 J

2. Total mass M = 40.0 g + 500 g = 540 g; M = 540 g = 0.540 kgFind vo of total mass M such that M rises h = 0.0450 m:Energy conservation: ½Mv2 + 0 = 0 + Mgh;

v = 0.939 m/s

3. Mass of box = 85/9.8 = 8.67 kg. Horizontally, resultant force, F = ma => 20 – fk = (8.67)(-0.90) => fk = 27.8NVertically, ΣF = 0 => n – 25 – 85 = 0 => n = 110 N.Frictional force, fk = µkn => µk = 0.253

4.

Potential energy of spring is converted to KE of 100 g mass:

Normal reaction, n = mg = (0.10)(9.8) = 0.98 N.

Work done against friction = loss in KE,

Frictional force,

5. (i) KEf = KEi + mgh½ mv 2 = ½ mu 2 + mghv2= u2 + 2gh = 36 + 2 × 9.8 × 3v = 9.74 ms–1

(ii) x = depression of the trampoline

.KEf + mgx = ½ kx2

½ × 60 × (9.74)2 + 60 × 9.8 x = ½ × 6.8 × 104 x2 34000x2 – 588x – 2846 = 0x = 0.298 m

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