*Notes to a video lecture on http://www.unizor.com*

__Problems on Mechanical Work__

*Problem A*

An object of mass

**is pushed up a plane inclined to horizon at angle**

*m***with constant acceleration**

*φ***to a height**

*a***against a friction with coefficient of kinetic friction**

*h***.**

*μ*(a) What is the amount of work necessary to achieve this goal?

(b) Consider the following real conditions of this experiment:

**m = 10kg**(kilograms)

**g = 9.8m/sec²**(meters per sec²)

**h = 0.5m**(meters)

**φ = 30°**(degrees)

**μ = 0.2**

**a = 0.1m/sec²**(meters per sec²)The work

**is measured in**

*W=F·S**N·m=kg·m²/sec²*units, called

*joule*and abbreviated as

**.**

*J*Calculate the work

**performed in this experiments in**

*W**joules*.

*Solution*

Let

**be the force that pushes the object up the slope,**

*F***is the object's weight,**

*P=m·g***is the force of resistance from the friction,**

*R***is the distance this object moves to reach the height**

*S***.**

*h*(a)

*F−P·sin(φ)−R = m·a*

*R = P·cos(φ)·μ*

*W = F·S*

*S = h/sin(φ)*

*F = P·sin(φ) + R + m·a =*

= m·[g·sin(φ) + g·cos(φ)·μ + a]= m·[g·sin(φ) + g·cos(φ)·μ + a]

*W = F·h/sin(φ) =*

=m·h[g+g·cot(φ)·μ+a/sin(φ)]=m·h[g+g·cot(φ)·μ+a/sin(φ)]

(b)

**W = 67J**(joules)*Problem B*

An object of mass

**is pushed by a constant force**

*m***down a slope of a plane inclined to horizon at angle**

*F***. Initially, it's at rest, the final speed is**

*φ***. There is a friction with coefficient of kinetic friction**

*V***.**

*μ*(a) What is the time

**from the beginning to the end of the object's motion?**

*t*(b) What is the distance

**this object moved until reaching the final speed**

*S***?**

*V*(c) What is the amount of work

**performed by this force?**

*W*(d) Consider the following real conditions of this experiment:

**F = 20N**(newtons)

**m = 10kg**(kilograms)

**g = 9.8m/sec²**(meters per sec²)

**V = 0.5m/sec**(meters per sec)

**φ = 5°**(degrees)

**μ = 0.7**Calculate the work

**performed in this experiments in**

*W**joules*, distance

**and time**

*S***of motion.**

*t**Solution*

**is the object's weight,**

*P=m·g***is the force of resistance from the friction,**

*R = P·cos(φ)·μ***is the distance this object moves to reach the speed**

*S***,**

*V***is the acceleration of this object.**

*a*Then from the Newton's Second Law

*F+P·sin(φ)−R = m·a*(we added

**because an object moves down a slope)**

*P·sin(φ)*Now we can find an acceleration of our object:

*F+m·g·sin(φ)−m·g·cos(φ)·μ =*

= m·a= m·a

*a = F/m + g·sin(φ) − g·cos(φ)·μ*Knowing acceleration

**and the correspondence between acceleration, final speed and time**

*a***, we can determine time:**

*V=a·t*(a)

*t = V/a*Now we can find the distance

(b)

*S = a·t²/2 = V·t/2*The work performed by the force

**equals to**

*F*(c)

*W = F·S*(d)

*a = 0.066(m/sec²)*

*t = 7.576(sec)*

*S = 1.894(m)*

*W = 37.879(joules))**Problem C*

An object of mass

**is dropped down from a certain height above a surface of some planet. At the moment it hits the ground its speed is**

*m***.**

*V*What is the work

**performed by the force of gravity?**

*W**Solution*

Let the acceleration of free falling on this planet is

**, the time of falling**

*a***and the height**

*t***.**

*S*Then

*F = m·a*

*V = a·t*

*t = V/a*

*S = a·t²/2 = V²/(2·a)*

*W = F·S = m·V²/2*Notice, the work depends only on mass and final speed - the results of

action, not on the height, nor acceleration of free falling, nor on

time.

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