Question

Force of 1:25 n acts on an object at rest after travelling a distance of 8 metre the speed of object in 10 M/S the mass of the object is as :
A) 1.25 kg
B) 2.5 kg
C )5 kg
D) 10kg
No spam!
Explain well plz!


:

Answer 1

$\huge \cal \underline \color{red}★A᭄ɴsᴡΣʀ࿐$

※Given Information,

• Force Applied (F) = 125 N
• Initial Velocity (u) = 0 m/s
• Final Velocity (v) = 10 m/s
• Displacement (s) = 8m
• From Work Energy Theorem,

※Net Work Done = Change in Kinetic Energy

Here, initial KE is inconsiquential as the object was at rest initially.

F × s = 1/2mv²

125 × 8 × 2 = 100m

100m = 250 × 8

100m = 2000

m = 20Kg

※Mass of the object is 20Kg.

________❤________

Answer 2

※Given Information,

※Given Information,Force Applied (F) = 125 N

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/s

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/s

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8m

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic Energy

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²125 × 8 × 2 = 100m

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²125 × 8 × 2 = 100m100m = 250 × 8

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²125 × 8 × 2 = 100m100m = 250 × 8100m = 2000

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²125 × 8 × 2 = 100m100m = 250 × 8100m = 2000m = 20Kg

※Given Information,Force Applied (F) = 125 NInitial Velocity (u) = 0 m/sFinal Velocity (v) = 10 m/sDisplacement (s) = 8mFrom Work Energy Theorem,※Net Work Done = Change in Kinetic EnergyHere, initial KE is inconsiquential as the object was at rest initially.F × s = 1/2mv²125 × 8 × 2 = 100m100m = 250 × 8100m = 2000m = 20Kg※Mass of the object is 20Kg.