Part (a): the work done by friction is
Part (b): the work done by rope is .
Part (c): the work done by gravitational force is .
Part (d): total work done is .
Further Explanation:
The rope is doing work against the gravity. The friction always acts against the direction of motion. Therefore, the friction is in direction of gravity.
Given:
Mass of victim is .
The inclination of plane is .
The coefficient of plane is .
The distance travelled on plane is .
Concept:
Equations for free body diagram of victim:
Force equation:
Normal force on sled is .
Friction force:
Substitute for
in above equation.
Here, is the friction force,
is the mass of victim,
is the gravitational acceleration,
is the coefficient of friction and
is the inclination of plane.
The gravitational force on the sled is .
Work equation:
(a)
Work done by friction:
Substitute for
in above equation.
Substitute for
,
for
,
for
,
for
and
for
in above equation.
Negative sign shows that friction is acting against the motion.
Thus, the work done by friction is
(b)
Work done by rope:
Substitute for
,
for
,
for
,
for
and
for
in above equation.
Thus, the work done by rope is .
(c)
Work done by gravitational force:
Substitute for
,
for
,
for
and
for
in above equation.
Thus, the work done by gravitational force is .
(d)
Total work done:
Substitute for
,
for
and
for
in above equation.
Thus, total work done is
Learn more:
1. Motion on a rough surface: brainly.com/question/7031524
2. Motion under the gravitational force: brainly.com/question/10934170
3. Principle of conservation of momentum: brainly.com/question/9484203
Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Ski patrol, rescue, sled, victim, mass, 90.0 kg, 60.0°, slope, coefficient, friction, snow, 0.100, work, 30.0 m, along, hill, distance and gravitational force
Answer:
The value is
Explanation:
From the question we are told that
The speed is
The radius of the earth is
Generally the circumfernce of the earth is mathematically evaluated as
=>
=>
Generally the time taken is mathematically represented as
Converting to days
=>
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s