You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps

5 0

4 years ago

A block lies on a frictionless floor. a force of 5 n pulls toward the east while a force of 4 n pulls toward the north. what is

sveticcg [70]

Given below the arrangement of loading on the larger boat by two tug boats.

F₁ = 5 N

F₂ = 4 N

Angle between them θ = 90⁰

Resultant between two vectors, $F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta }$

Substituting

$F = \sqrt{5^2+4^2+2*5*4*cos 90} \\ \\ = 6.403 N$

So magnitude of the net force on the block = 6.403 N

3 0

4 years ago

A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000

34kurt

Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N

4 0

3 years ago