Mekanika Practice Questions
UN (Indonesia) · UN Fisika · 138 free MCQs with instant results and detailed explanations.
138
Total
42
Easy
72
Medium
24
Hard
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Sample Questions from Mekanika
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
Two objects are in equilibrium. If the force acting on one object is 10 N to the right, what is the force acting on the other object?
A.
10 N to the left
B.
5 N to the right
C.
10 N to the right
D.
0 N
Show Answer & Explanation
Correct Answer: A
In equilibrium, the net force acting on an object must be zero. Therefore, if one object has a force of 10 N to the right, the other must exert a force of 10 N to the left to balance it.
Q2
Easy
An object is moving with a constant velocity. Which of the following statements is true?
A.
The net force acting on the object is zero.
B.
The object is experiencing a constant acceleration.
C.
The object will eventually stop.
D.
The speed of the object is increasing.
Show Answer & Explanation
Correct Answer: A
When an object moves with a constant velocity, it means that there is no change in speed or direction, which implies that the net force acting on the object is zero according to Newton's first law of motion.
Q3
Easy
A car accelerates from rest at a rate of 2 m/sยฒ. How far does it travel in 5 seconds?
A.
25 meters
B.
20 meters
C.
15 meters
D.
10 meters
Show Answer & Explanation
Correct Answer: A
Using the formula for distance under constant acceleration: distance = 0.5 ร acceleration ร timeยฒ. Here, distance = 0.5 ร 2 m/sยฒ ร (5 s)ยฒ = 25 meters.
Q4
Medium
A block of mass 10 kg is sliding down a frictionless incline which makes an angle of 30ยฐ with the horizontal. What is the acceleration of the block?
A.
4.9 m/sยฒ
B.
9.8 m/sยฒ
C.
3.4 m/sยฒ
D.
5.0 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration can be calculated using the formula a = g * sin(ฮธ). Here, g = 9.8 m/sยฒ and ฮธ = 30ยฐ. Thus, a = 9.8 * 0.5 = 4.9 m/sยฒ.
Q5
Medium
A spring has a spring constant of 200 N/m. If it is compressed by 0.1 m, what is the potential energy stored in the spring?
A.
1 J
B.
2 J
C.
0.5 J
D.
10 J
Show Answer & Explanation
Correct Answer: A
The potential energy stored in a spring is given by PE = (1/2)kxยฒ. Here, k = 200 N/m and x = 0.1 m. Therefore, PE = 0.5 * 200 * (0.1)ยฒ = 1 J.
Q6
Medium
Which of the following quantities is conserved in a perfectly elastic collision?
A.
Momentum only
B.
Kinetic energy only
C.
Both momentum and kinetic energy
D.
Neither momentum nor kinetic energy
Show Answer & Explanation
Correct Answer: C
In a perfectly elastic collision, both momentum and kinetic energy are conserved. This distinguishes it from inelastic collisions where kinetic energy is not conserved.
Q7
Medium
A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. What is the acceleration of the car?
A.
4 m/sยฒ
B.
2 m/sยฒ
C.
5 m/sยฒ
D.
10 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration can be calculated using the formula a = (v - u) / t. Here, u = 0 m/s, v = 20 m/s, and t = 5 s, so a = (20 - 0) / 5 = 4 m/sยฒ.
Q8
Hard
A block of mass 5 kg is sliding down a frictionless incline of angle 30ยฐ with respect to the horizontal. What is the acceleration of the block down the incline?
A.
4.9 m/sยฒ
B.
9.8 m/sยฒ
C.
3.9 m/sยฒ
D.
6.9 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration of an object on an incline is given by the formula a = g * sin(ฮธ). Here, g = 9.8 m/sยฒ and ฮธ = 30ยฐ. Thus, a = 9.8 * sin(30ยฐ) = 9.8 * 0.5 = 4.9 m/sยฒ.
Q9
Hard
A block of mass 5 kg is sliding down a frictionless incline of 30 degrees. Calculate the acceleration of the block down the incline. (Use g = 9.8 m/sยฒ)
A.
4.9 m/sยฒ
B.
8.5 m/sยฒ
C.
9.8 m/sยฒ
D.
5.0 m/sยฒ
Show Answer & Explanation
Correct Answer: A
The acceleration of an object on an incline is given by g * sin(theta). Here, theta = 30 degrees, so sin(30) = 0.5. Therefore, the acceleration a = 9.8 * 0.5 = 4.9 m/sยฒ.
Q10
Hard
A 2 kg object is attached to a spring with a spring constant of 300 N/m. If the object is pulled 0.1 m from its equilibrium position and released, what is the maximum speed of the object as it passes through the equilibrium position?
A.
1.73 m/s
B.
3.46 m/s
C.
2.00 m/s
D.
4.00 m/s
Show Answer & Explanation
Correct Answer: A
The maximum speed can be calculated using the conservation of energy principle. The potential energy in the spring when stretched is converted to kinetic energy at the equilibrium position. PE = 1/2 k x^2 = 0.5 * 300 * (0.1)^2 = 1.5 J. Set this equal to KE = 1/2 m v^2. Solving for v gives v = sqrt(3/m) = sqrt(3/2) = 1.73 m/s.
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Mekanika โ UN (Indonesia) UN Fisika Practice Questions Online
This page contains 138 practice MCQs for the chapter Mekanika in UN (Indonesia) UN Fisika. The questions are organized by difficulty โ 42 easy, 72 medium, 24 hard โ so you can choose the right level for your preparation.
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