Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by

where k is the elastic constant of the spring and

is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:

where the negative sign is given by the fact that

points in the opposite direction of the displacement of the cart, and where

therefore, the work done by the weight is
Answer:84.672 joules.
Explanation:
1) Data:
m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²
2) Physical principle
Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.
3) Calculations:
The gravitational potential energy, PE, is equal to m × g × h
So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.
And that is the kinetic energy that the dog needs.
Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:

Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:

The minimum stopping distance is 141.78 ft.
Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.