Question

❀A block of mass 500g is pulled from rest on a horizontal frictionless bench by a steady force F and travels 8m in 2s find
A. acceleration
B. the value of F​​

Answer 1

Equations:

F=m*a

d=vi*t+1/2*a*t²

Given Variables

m=.5kg

vi=0m/s

d=8m

t=2s

8=0m/s*2s+1/2*a*2s²

a=4m/s²

F=.5kg*4m/s

F=2N

Answer 2

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Given :

Initial velocity = zero

Mass of block = 500g

Applied force = F

Distance travelled in 2s = 8 m

To Find :

Acceleration

Value of applied force

Solution :

❖ Assuming that body has constant acceleration throughout the motion..

Acceleration of block can be easily calculated by applying equation of kinematics.

Second equation of kinematics :

d = ut + 1/2 at²

» d denotes distance

» u denotes initial velocity

» t denotes time

» a denotes acceleration

By substituting the given values;

➙ d = ut + 1/2 at²

➙ 8 = (0 × 2) + 1/2 a (2)²

➙ 8 = 0 + 4a/2

➙ 8 = 2a

➙ a = 8/2

➙ a = 4 m/s²

♦ As per newton's second law of motion, force is measured as the product of mass and acceleration.

Mathematically, F = m a

[Mass (m) = 500g = 0.5kg]

➙ F = 0.5 × 4

➙ F = 2 N

hope it helps uh :)