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
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s