Answer:
1.8 m/s
Explanation:
Draw a free body diagram of the block. There are four forces:
Normal force Fn up.
Weight force mg down.
Applied force F to the east.
Friction force Fn μ to the west.
Sum the forces in the y direction:
∑F = ma
Fn − mg = 0
Fn = mg
Sum the forces in the x direction:
F − Fn μ = ma
F − mg μ = ma
a = (F − mg μ) / m
a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg
a = 0.53 m/s²
Given:
Δx = 3 m
v₀ = 0 m/s
a = 0.53 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)
v = 1.8 m/s
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
