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In this lesson, we will learn how to calculate the work done by a constant force vector acting on a body over a displacement vector using the dot product.

Q1:

A particle moves in a plane in which β π and β π are perpendicular unit vectors. A force, β πΉ = οΊ 9 β π + β π ο N acts on the particle. The particle moves from the origin to the point with position vector οΊ β 9 β π + 6 β π ο m. Find the work done by the force.

Q2:

A force β πΉ = οΊ π β π β 9 β π ο N acts on a particle, causing a displacement β π = ο β 5 β π + ( π + 6 ) β π ο c m . If the work done by the force is 0.02 J, what is the value of π ?

Q3:

A particle moved from point π΄ ( β 7 , β 1 ) to point π΅ ( β 4 , 6 ) along a straight line under the action of the force β πΉ = π β π + π β π . During this stage of motion, the work done by the force was 106 units of work. The particle then moved from π΅ to another point πΆ ( β 8 , β 3 ) under the effect of the same force. During this stage of motion, the work done by the force was β 1 3 8 units of work. Determine the two constants π and π .

Q4:

A particle moved from point π΄ ( β 2 , β 2 ) to point π΅ ( 6 , 1 0 ) along a straight line under the action of the force β πΉ = π β π β 6 β π acting in the opposite direction to the displacement ο π΄ π΅ . Find the work done by the force β πΉ .

Q5:

A particle moved on a plane from the point π΄ ( β 8 , 6 ) to the point π΅ ( 2 , 5 ) under the action of a force of 17 N whose line of action made an angle of π with the π₯ -axis where s i n π = 8 1 7 . Find the work done by this force over the displacement ο π΄ π΅ .

Q6:

A particle moved from point π΄ ( 7 , β 3 ) to point π΅ ( β 9 , 2 ) along a straight line under the action of a force β πΉ of magnitude 8 β 1 0 N acting in same direction as the vector β π = β 3 β π β β π . Calculate the work done by the force, given that the magnitude of the displacement is measured in metres.

Q7:

A body of mass 3 kg is moving under the action of a force β πΉ , such that its displacement β π ( π‘ ) = οΉ 5 π‘ ο β π + ( 7 π‘ ) β π 2 . Find the work done by this force in the first 6 seconds of its motion, given that the displacement is measured in metres, the force in newtons, and the time π‘ in seconds.

Q8:

The displacement of a particle of mass 30 g is given as a function of time by the relation , where is a constant unit vector, is measured in centimeters, and in seconds. Given that the particle started its motion at , find the force acting on the particle and the work done by this force during the first 7 seconds of motion.

Q9:

A particle moves in a plane in which β π and β π are perpendicular unit vectors. Its displacement from the origin at time t seconds is given by β π = ο οΉ 2 π‘ + 7 ο β π + ( π‘ + 7 ) β π ο 2 m and it is acted on by a force β πΉ = οΊ 6 β π + 3 β π ο N . how much work does the force do between π‘ = 2 s and π‘ = 3 s ?

Q10:

The position vector of a particle of mass 3 kg moving under the action of a force is given as a function of time π‘ by the relation r i j ( π‘ ) = ( β 4 π‘ β 1 0 ) + ( 3 π‘ + 5 ) ο¨ ο¨ , where i and j are two perpendicular unit vectors. Calculate the work done by the force between π‘ = 3 to π‘ = 4 .

Q11:

A body of mass 2 kg is moving under the action of three forces, β πΉ 1 , β πΉ 2 , and β πΉ 3 , where β πΉ = π β π β 3 β π 1 , β πΉ = β 4 β π + 3 β π 2 , and β πΉ = β 1 0 β π + π β π 3 , and β π and β π are two perpendicular unit vectors, π and π are constants, and each force is measured in newtons. The displacement of the body is expressed by the relation β π ( π‘ ) = οΉ 4 π‘ ο β π + οΉ 3 π‘ β 8 π‘ ο β π 2 2 , where the displacement is measured in metres, and the time π‘ is in seconds. Determine the work done by the resultant of the forces in the first 6 seconds of motion.

Q12:

A body moves in a plane in which β π and β π are perpendicular unit vectors. At time π‘ seconds, its position vector is given by β π = ο ( β 2 π‘ + 8 ) β π + οΉ β π‘ + 1 0 π‘ ο β π ο 2 m . Given that from π‘ = 5 s to π‘ = 8 s , the change in the bodyβs kinetic energy was 414 J, find the bodyβs mass.

Q13:

A force F i j = ( β 4 β 9 ) N is acting on a particle whose position vector as a function of time is given by r i j ( π‘ ) = οΉ ( β 9 π‘ β 8 ) + οΉ β 3 π‘ + 2 ο ο ο¨ m. Calculate the work π done by the force F between π‘ = 3 and π‘ = 8 s e c o n d s .

Q14:

An object moves 10 m in the direction of . There are two forces acting on this object: N and N. Find the total work done on the object by the two forces.

Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

Q15:

Q16:

An object moves 20 metres in the direction of . There are two forces acting on this object: N and N. Find the total work done on the object by the two forces.

Q17:

A body moves under the force F i j = 6 β 9 from point π΄ ( β 2 , 8 ) to point π΅ ( 1 , β 7 ) . Determine the work done by the force, where the displacement is measured in metres and the force in newtons.

Q18:

A force of β πΉ = 6 β π + 4 β π n e w t o n s acts on a body and moves it from point π΄ ( β 9 , β 2 , β 3 ) to point π΅ ( 8 , β 7 , β 1 ) . Determine the work done by the force β πΉ , where the displacement is measured in metres.

Q19:

A body moves from point π΄ ( 1 , 2 , 5 ) to π΅ ( 5 , 5 , β 4 ) under a force β πΉ . Determine the work done if the force has magnitude β 5 7 newtons and direction cosines οΏ 5 β 5 7 5 7 , 4 β 5 7 5 7 , 4 β 5 7 5 7 ο , taking the displacement as measured in metres.

Q20:

A particle is moving in a straight line under the action of the force β πΉ = β 8 β π β 3 β π from point π΄ ( 8 , 7 ) to point π΅ ( 8 , β 5 ) . Find the work done π by the force β πΉ .

Q21:

A body is displaced by β π = β 6 β π β 5 β π β 7 β π metres with a force β πΉ = 8 β π β 1 0 β π newtons. What is the work done?

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