A block of ice with a mass of 2.50 kg is moving on a frictionless, horizontal surface. At time t = 0, the block is moving to the right with a velocity of magnitude 8.00 m/s. Calculate the velocity of the block after a force of 7.00 N directed to the left has been applied for 5.00 s.

Respuesta :

Answer:

The velocity of the block is 22 m/s.

Explanation:

Given that,

Mass = 2.50 kg

Velocity = 8 .00 m/s

Force = 7.00 N

Time t = 5.00

We need to calculate the change in velocity it means acceleration

Using newton's law

[tex]F = ma[/tex]

Where,

m = mass

a = acceleration

Put the value into the formula

[tex]a=\dfrac{F}{m}[/tex]

[tex]a = \dfrac{7.00}{2.50}[/tex]

[tex]a= 2.8m/s^2[/tex]

We need to calculate the velocity of the block

Using equation of motion

[tex]v = u+at[/tex]

Where,

v = final velocity

u = initial velocity

a = acceleration

t =time

Put the value in the equation

[tex]v= 8.00+2.8\times5.00[/tex]

[tex]v=22\ m/s[/tex]

Hence, The velocity of the block is 22 m/s.