3.6×0.48×2.50 upon 0.12×0.99×0.5
Answers
Step-by-step explanation:
Given
A block of mass 500 kg starts from rest and moves with speed of 10 m/s within 5 seconds
To Find
Force acting on the block
Solution
We know that
\rm \longrightarrow F=m \times a⟶F=m×a
\rm \longrightarrow a=\bigg(\dfrac{v-u}{t}\bigg)⟶a=(
t
v−u
)
Hence
By equating 'a'
We get
\rm \longrightarrow F=m \times \bigg(\dfrac{v-u}{t}\bigg)⟶F=m×(
t
v−u
)
Here
F = Force
m = Mass
a = Acceleration
u = Initial velocity
v = Final velocity
t = Time
Units
F = Newton (N)
m = kilogram (kg)
a = metre/square second (m/s²)
u = metre/second (m/s)
v = metre/second (m/s)
t = second (s)
According to the question
We are asked to find the force acting on the block
Therefore
We must find "F"
Given that
A block of mass 500 kg starts from rest and moves with speed of 10 m/s within 5 seconds
Hence
m = 500 kg
u = 0 m/s [starts from rest]
v = 10 m/s
t = 5 s
Substituting the values
We get
\rm \longrightarrow F=500 \times \bigg(\dfrac{10-0}{5}\bigg) \ N⟶F=500×(
5
10−0
) N
\rm \longrightarrow F=500 \times \bigg(\dfrac{10}{5}\bigg) \ N⟶F=500×(
5
10
) N
\rm \longrightarrow F=100 \times 10 \ N⟶F=100×10 N
Therefore
\rm \longrightarrow F=1000 \ N⟶F=1000 N
Hence
Force acting on the block = 1000 N
See the image for answer.
