Physics question. Attempted the problem but got wrong answer.

Question

A 2kg block is placed at rest on a

friction less incline that is angled 25 degrees. It is

released and slides 30.0 cm (distance s)

along the incline before reaching the

floor. The floor exerts a kinetic friction

force f of 1 N on the block. How far does

the block slide (distance x) on the flat

section (different from s) before it stops?

Answer 1

You need to first find the speed of the block when it reaches the floor, and then use kinematics to get the stopping distance on the floor. Since it slides down the plane without friction, energy is conserved, so just set the potential energy at the top equal to the kinetic energy at the bottom and solve for the speed.

mgh = mv₁²/2

h = initial height = s sinθ

gs sinθ = v₁²/2

v₁ =√(2gs sinθ)

Plug in

g = 9.8m/s²

s = 0.30 m

θ = 25°

and get v₁. Now we are ready to get the horizontal sliding distance x, from kinematics.

v₂² - v₁² = 2ax

v₂ = 0 (comes to a full stop)

You have v₁ from the steps above.

a = f/m = -1N / (2kg) = ? m/s²

Note the acceleration is negative, since the block is slowing down. Put it all together and get

-v₁² = 2ax

x = -v₁² / (2a)

Keep the negative sign for a, and x will come out as a positive distance. You should be able to finish from here.

Answer 2

First determine the velocity of the object at the bottom of the incline before it slides against the friction.

using the equation (V_f^2 - V_o^2) = 2*a*deltaX

where V_0 (initial velocity) = 0 m/s

a = acceleration which will solve for below

deltaX is distance traveled = 30 cm

V_f is velocity at bottom of incline (what we want to find)

gravity is the only force acting on the object on the incline of 25 degrees so the acceleration due to gravity parallel to the incline is g*sin(25) where g is 9.8 m/s^2

plugging into equation

V_f² - 0² = 2*(9.8m/s^2)(sin(25^o)*0.3 m

V_f = 1.58 m/s

now to determine how far the object will slide we must determine how the friction force decelerates the block. The friction force is 1 N and because F= ma and m = 2 kg, the deceleration due to friction is 0.5 m/s^2.

We can then use the same equation used in the previous part with the following terms:

V_0 = 1.58 m/s (velocity of object at bottom of incline)

V_f = 0 m/s (because the object ends at rest)

a = -0.5m/s^2

and solve for deltaX

deltaX = 2.4964 m