Force Practice Questions
KCSE (Kenya) · KCSE Physics · 148 free MCQs with instant results and detailed explanations.
148
Total
55
Easy
70
Medium
23
Hard
Start Practicing Force
Take a timed quiz or customize your practice session
Sample Questions from Force
Here are 10 sample questions. Start a quiz to get randomized questions with scoring.
Q1
Easy
A 5 kg object is pushed with a force of 20 N. What is its acceleration?
A.
2 m/sĀ²
B.
4 m/sĀ²
C.
5 m/sĀ²
D.
10 m/sĀ²
Show Answer & Explanation
Correct Answer: B
Using Newton's second law, F = ma, we can solve for acceleration (a). Rearranging gives a = F/m. Thus, a = 20 N / 5 kg = 4 m/sĀ².
Q2
Easy
If two forces of 10 N and 15 N are acting in the same direction on an object, what is the resultant force?
A.
5 N
B.
25 N
C.
15 N
D.
10 N
Show Answer & Explanation
Correct Answer: B
When two forces act in the same direction, the resultant force is the sum of the two forces. Therefore, 10 N + 15 N = 25 N.
Q3
Easy
What is the force required to accelerate a 10 kg mass at 2 m/sĀ²?
A.
20 N
B.
15 N
C.
5 N
D.
25 N
Show Answer & Explanation
Correct Answer: A
According to Newton's second law, Force (F) equals mass (m) multiplied by acceleration (a). Here, F = 10 kg Ć 2 m/sĀ² = 20 N.
Q4
Medium
A car accelerates from rest at a rate of 2 m/sĀ² for 10 seconds. What is the final velocity of the car?
A.
20 m/s
B.
15 m/s
C.
10 m/s
D.
25 m/s
Show Answer & Explanation
Correct Answer: A
The final velocity can be calculated using the formula v = u + at. Here, u = 0 m/s (initial velocity), a = 2 m/sĀ² (acceleration), and t = 10 s (time). Thus, v = 0 + (2 * 10) = 20 m/s.
Q5
Medium
A cyclist traveling at 10 m/s applies brakes and comes to a stop in 5 seconds. What is the average deceleration of the cyclist during this time?
A.
2 m/sĀ²
B.
1 m/sĀ²
C.
3 m/sĀ²
D.
4 m/sĀ²
Show Answer & Explanation
Correct Answer: A
Deceleration is calculated as the change in velocity divided by the time taken. Here, initial velocity = 10 m/s, final velocity = 0 m/s, and time = 5 s. So, deceleration = (10 - 0) / 5 = 2 m/sĀ².
Q6
Medium
If an object of mass 10 kg is acted upon by a net force of 50 N, what is the object's acceleration?
A.
5 m/sĀ²
B.
10 m/sĀ²
C.
15 m/sĀ²
D.
20 m/sĀ²
Show Answer & Explanation
Correct Answer: A
Using Newton's second law, F = ma, we can solve for acceleration a = F/m. Here, F = 50 N and m = 10 kg. Thus, a = 50/10 = 5 m/sĀ².
Q7
Medium
A 2 kg object is in free fall near the surface of the Earth. What is the force acting on the object due to gravity?
A.
9.8 N
B.
2 N
C.
19.6 N
D.
4.9 N
Show Answer & Explanation
Correct Answer: C
The force acting on an object due to gravity can be calculated using F = mg. Given m = 2 kg and g = 9.8 m/sĀ² (acceleration due to gravity), the force is F = 2 * 9.8 = 19.6 N.
Q8
Hard
A car of mass 1200 kg is moving at a speed of 20 m/s when it suddenly applies brakes, producing a constant deceleration of 5 m/sĀ². How far does the car travel before coming to a complete stop?
A.
40 m
B.
60 m
C.
80 m
D.
100 m
Show Answer & Explanation
Correct Answer: C
The distance traveled before the car stops can be calculated using the formula: vĀ² = uĀ² + 2as. Here, v = 0 m/s (final velocity), u = 20 m/s (initial velocity), and a = -5 m/sĀ² (deceleration). Plugging in the values: 0 = (20)Ā² + 2(-5)s gives s = 80 m. Thus, option C is correct.
Q9
Hard
A block of mass 5 kg is placed on a frictionless surface and is pulled by a force of 20 N. What is the acceleration of the block?
A.
4 m/sĀ²
B.
2 m/sĀ²
C.
5 m/sĀ²
D.
10 m/sĀ²
Show Answer & Explanation
Correct Answer: A
Using Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration. Rearranging gives a = F/m. Substituting the values: a = 20 N / 5 kg = 4 m/sĀ². Therefore, the acceleration of the block is 4 m/sĀ².
Q10
Hard
A car of mass 1200 kg is moving at a speed of 25 m/s when the driver applies brakes, providing a deceleration of 5 m/sĀ². How much distance will the car cover before coming to a stop?
A.
62.5 m
B.
75 m
C.
125 m
D.
150 m
Show Answer & Explanation
Correct Answer: A
Using the equation vĀ² = uĀ² + 2as, where v = final velocity (0 m/s), u = initial velocity (25 m/s), a = acceleration (-5 m/sĀ²), we find: 0 = 25Ā² + 2*(-5)*s. Solving gives s = 62.5 m.
Showing 10 of 148 questions. Start a quiz to practice all questions with scoring and timer.
Practice All 148 Questions →Other KCSE (Kenya) Subjects
Force ā KCSE (Kenya) KCSE Physics Practice Questions Online
This page contains 148 practice MCQs for the chapter Force in KCSE (Kenya) KCSE Physics. The questions are organized by difficulty ā 55 easy, 70 medium, 23 hard ā so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.