A block of mass 5kg is pulled with a constant force of 25N with the help of a
string. The coefficient of friction between the block and the ground is 0.2. Pulley is
massless and string is massless & inextensible. When the string makes an angle of
37" with the horizontal, the acceleration of the block is
(A) 2
(B4m/s
(C)5m/s
(D) 2.6m
Answers
Answer:
the answer = 2m/s
let acceleration is friction M=0.2
F=25N
ma=Fcos 37-mgM
therefore;a=2m/s
The acceleration of the block is (a) 2 m/s²
Given: Mass of block = 5 kg, Magnitude of force = 25 N,
The coefficient of friction between the block and the ground is 0.2.
The string makes an angle of 37° with the horizontal.
To Find: The acceleration of the block.
Solution:
- The question can be solved by using the formula,
∑ F = m × a [ F = force, m = mass, a = acceleration ]
- We know that frictional force always acts in the direction opposite to the direction of motion.
Coming to the numerical,
m = 5 kg, F = 25 N, μ = 0.2,
Component of force acting along the direction of force = F cos 37°
= 25 × 4 / 5
= 20 N
Force of friction = μmg
= 0.2 × 5 × 10
= 10 N
According to the question,
∑ F = m × a
⇒ F cos 37° - μmg = m × a
⇒ 20 - 10 = 5 × a
⇒ a = 10 / 5
⇒ a = 2 m/s²
Hence, the acceleration of the block is (a) 2 m/s².
#SPJ3