Answer:
Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
v2f=v2i−2⋅a⋅d
Isolate d on one side of the equation and solve by plugging your values
d=v2i−v2f2a
d=(15.02−10.02)m2s−22⋅2.0ms−2
d=31.3 m
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
vf=vi−a⋅t, which will get you
t=vi−vfa
t=(15.0−10.0)ms2.0ms2=2.5 s
Answer: 6250 joules
Explanation:
The work needed to lift an object of mass M by a height H is equal to:
w = M*g*H
where h = 10m/s^2
then the total work that he did is equal to the sum of the work for every stone:
W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)
= (100kg + 120kg + 140kg + 160kg + 180kg)*g*H
= (500kg)*g*H
and now we can repalce g by 10m/s^2 and H by 125cm
But you can notice that we have two different units of distance, so knowing that 100cm = 1m
we can write H = 125cm = (125/100) m = 1.25 m
Then we have:
H = 500kg*10m/s^2*1.25m = 6250 J
Answer:
m = 0.51[kg]
Explanation:
Potential energy is defined as the product of mass by gravity by height.
where:
Epot = potential energy = 15 [J]
m = mass [kg]
g = gravity acceleration = 9.8 [m/s²]
h = elevation = 3 [m]
Now replacing:
![E_{pot}=m*g*h\\15=m*9.8*3\\m = 0.51[kg]](https://tex.z-dn.net/?f=E_%7Bpot%7D%3Dm%2Ag%2Ah%5C%5C15%3Dm%2A9.8%2A3%5C%5Cm%20%3D%200.51%5Bkg%5D)
