A horizontal conveyor belt moves with a constant velocity V. A small block is projected
with a velocity of 6 m/s on it in a direction opposite to the direction of motion of the belt.
The block comes to rest relative to the belt in a time 4s. j = 0.3, 9= 10 m/s2. Find V.
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Answer:
6/4 m/s^2 or 1.5 m/s^2
Explanation:
See the attached file
Simple question of motion in one dimension
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Given:
A horizontal conveyor belt moves with a constant velocity V.
To find:
Find V.
Solution:
From given, we have the data as follows.
A small block is projected with a velocity of 6 m/s on it in a direction opposite to the direction of motion of the belt.
⇒ V_{b/c} = V_{6} + V
The block comes to rest relative to the belt in a time 4s. j = 0.3, 9= 10 m/s2.
The frictional force is,
f = μmg
f = 0.3 × m × 10
f = 3m
The acceleration is,
a = F/m
a = 3m/m
a = 3 m/s²
The law of motion equation.
v = u + at
6 + V = 0 + 3 × t
6 + V = 3 × 4
6 + V = 12
V = 12 - 6 = 6
∴ V = 6 m/s
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