Explanation:
It is given that,
Mass of the truck, m = 2000 kg
Initial velocity of the truck, u = 34 km/h = 9.44 m/s
Final velocity of the truck, v = 58 km/h = 16.11 m/s
(a) Change in truck's kinetic energy,
(b) Change in momentum of the truck,
Hence, this is the required solution.
Answer:
Explanation:
Frictional force acting on incline = μ mg cosθ
μ is coefficient of friction , m is mass of object , θ is incline
= .09 x m x 9.8 x cos 28
= .78 m
work done by friction
= frictional force x displacement
= - .78m x 100
= - 78m
Potential energy of sli at height
= mgh
= m x 9.8 x 100 sin 28
= 460.08 m
net energy at the base
= 460.08m - 78 m
= 382.08 m
This will be in the form of kinetic energy .
1/2 m v² = 382.08 m
.5 x v² = 382.08
v = 27.64 m/s
After that it travels on plane surface .
Let the distance travelled be d
work done by frictional force
= μ mg x d
= .09 x m x 9.8 x d
This will be equal to kinetic energy at the base
.09 x m x 9.8 x d = 382.08 m
d = 433.2 meter .
Answer:
False
Explanation:
The potential energy of a body increases by increasing the distance or elevation from a reference level (ground). This energy can be easily calculated by means of the following equation.
where:
m = mass [kg]
g = gravity acceleration [m/s²]
h = elevation [m]
As we can see the potential energy is proportional to height, that is, as the body increases its elevation, its potential energy increases.
Explanation:
The given data is as follows.
mass (m) = 5 kg
Height of tower = 15 m
u = 7 m/s
air resistance = 610 v
(a) Now, differential equation for the given mass which is thrown vertically upwards is as follows.
= F
-bv = Fr
Here, mg is downwards due to the force of gravity.
= 0
Hence, the differential equation required to solve the problem is as follows.
= 0
(b) When final velocity of the object is equal to zero then the object will reach towards its maximum height and it will start to fall downwards.
F =
= 0
Therefore, the object reach its maximum height at v = 0.