Question

need help!! 1.A car has a mass of 1.72x103kg.
If the force acting on the car is 6.44x103
N to the east,what is the car’s acceleration?
Answer in units of m/s2
2.What is the acceleration of a 56 kg block of
cement when pulled sideways with a net force
of 526 N?
Answer in units of m/s2
6.A 1553.8 kg car is traveling at 26.6 m/s
when the driver takes his foot off the gas
pedal. It takes 4.9 s for the car to slow down
to 20 m/s.
How large is the net force slowing the car?
Answer in units of N

Answer 1

F=ma

Answer 2

m=1720kg, F=6440N=6440kgms^-2, so what is a?

Answer 3

same thing for the rest

Answer 4

1. You need to use Newton's second law, F=m*a, for this problem.
In order to solve for acceleration, divide force by the mass.
a=F/m
Now, plug in your variables. You're given Force and mass, so this is just putting numbers into an equation.
a=(6.44*10^3)/(1.72*10^3)
a=3.744 m/s^2

2. Same equation as the first one: F=m*a
In order to solve for acceleration, divide force by the mass.
a=F/m
You're given Force and mass, so put those numbers into the equation.
a=526/56
a=9.393 m/s^2

6. First, you need to solve for acceleration, so use the equation Vf=Vi+at
(Vf is final velocity, which would be 20 m/s, and Vi is initial velocity, which would be 26.6 m/s)
So to solve for acceleration, subtract Vi and then divide by time.
Vf=Vi+at
Vf-Vi=at
(Vf-Vi)/t=a or a=(Vf-Vi)/t
Now you can plug in Vf, Vi, and t to find the acceleration.
a=(Vf-Vi)/t
a=(20-26.6)/4.9
a=-6.6/4.9
a=-1.347
Acceleration is negative because you're slowing the car down.
Now, you can use F=ma to solve.
F= 1553.8*-1.347
F=-2092.873 N

Hope that helps!

Answer 5

thank you Poeticalto