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In this lesson, we will learn how to apply Newtonβs second law when the forces acting on a body and the motion caused by them are represented in the vector notation.

Q1:

If a body of mass 1 kg moves under the action of forces β πΉ = ο» β π + 8 β π β 5 β π ο 1 N and β πΉ = ο» 2 β π β 7 β π + 8 β π ο 2 N , what is its acceleration?

Q2:

A body of mass 11 kg is moving such that the horizontal and vertical components of its velocity are given by π£ = 4 π₯ and π£ = β 9 . 8 π‘ + 1 2 π¦ where π£ π₯ and π£ π¦ are measured in metres per second. Find the force β πΉ , in newtons, that is acting on the body during its motion and the bodyβs initial speed π£ 0 .

Q3:

A body of mass 3 units was moving under the action of two coplanar forces β πΉ 1 and β πΉ 2 such that β πΉ = π β π + 4 β π 1 and β πΉ = β 4 β π + π β π 2 , where β π and β π are two perpendicular unit vectors. Given that the acceleration of the body is 2 β π β 4 β π , find the values of the constants π and π .

Q4:

A particle of mass π kg is moving under the action of two forces: β πΉ = 8 π β π + 6 π β π 1 and β πΉ = 4 π β π 2 , where β π and β π are two perpendicular unit vectors. Find the acceleration β π of the particle and its magnitude β β β π β β in metres per second squared.

Q5:

If the forces β πΉ = ο» π₯ β π + π¦ β π + π§ β π ο 1 N and β πΉ = ο» β 5 β π β 6 β π β 3 β π ο 2 N acting on a body of mass 6 kg, cause an acceleration β π = ο» 5 β π + 2 β π β 4 β π ο / m s 2 , what are the values of π₯ , π¦ , and π§ ?

Q6:

Given that the motion of a body of mass 2 kg is represented by the relation β π ( π‘ ) = οΉ 6 π‘ + 1 5 π‘ + 2 ο β π 2 , where β π is a constant unit vector, β π is measured in metres, and π‘ is measured in seconds, determine the magnitude of the force acting on the body.

Q7:

A body of unit mass was moving under the effect of a force β πΉ = π β π + π β π , where β π and β π are two orthogonal unit vectors. If the displacement vector of the body at time π‘ is given by β π ( π‘ ) = ( 9 π‘ ) β π + ( π‘ + 3 ) β π 2 2 , find π and π .

Q8:

A particle of unit mass was moving under the effect of three forces: β πΉ = π β π 1 , β πΉ = β β π 2 , and β πΉ = 2 β π + π β π 3 , where β π and β π are two perpendicular unit vectors and π and π are constants. If the displacement vector of the particle as a function of the time is given by β π ( π‘ ) = 6 β π + ( β 4 π‘ + 4 π‘ ) β π 2 , find the values of π and π .

Q9:

A body of mass 9 g was moving on a plane under the effect of the force β πΉ = οΊ β β π β 1 0 β π ο dynes. Given that the position vector of the body is given by the relation β π ( π‘ ) = οΊ οΉ π π‘ + 7 ο β π + οΉ π π‘ + 6 π‘ ο β π ο 2 2 c m , determine π and π .

Q10:

A body of mass 7 kg moves under the action of three forces, β πΉ = οΊ π β π + 3 β π ο 1 N , β πΉ = οΊ 6 β π β 6 β π ο 2 N , and β πΉ = οΊ 6 β π + π β π ο 3 N . Given that the displacement of the particle at time π‘ seconds is β π = ο οΉ π‘ + 6 ο β π + οΉ 5 π‘ + 5 ο β π ο 2 2 m , determine the values of π and π .

Q11:

A particle of unit mass is moving such that its velocity at a given time π‘ is represented by β π£ ( π‘ ) = οΉ 8 π π‘ + 5 π π‘ ο β π 2 , where β π is a constant unit vector. Given that the force acting on the particle at time π‘ is β πΉ ( π‘ ) = ( 1 0 π‘ + 4 ) β π , find π and π .

Q12:

A particle of unit mass is moving along a certain path, where its velocity at time π‘ is given by the relation β π£ = οΉ π π‘ + π π‘ ο β π 2 , where β π is a constant unit vector. Given that the force acting on the particle is constant and given by the relation β πΉ = 9 1 β π , determine the values of the constants π and π .

Q13:

A body of mass 250 g moves under the action of a force, β πΉ newtons. Given that the body starts from rest at the origin, and β πΉ = ( 9 π‘ + 3 ) β π + 9 π‘ β π , where β π and β π are perpendicular unit vectors, find the displacement in terms of π‘ .

Q14:

A particle of mass 5 kg was in motion. The components of its velocity in the horizontal and vertical directions were π£ = 3 / π₯ m s and π£ = ( β 4 . 7 π‘ + 1 4 ) / π¦ m s , respectively. Determine the magnitude, π£ 0 , and direction, π , of its initial velocity and the force β πΉ acting on it.

Q15:

A body of mass π is moving under the action of a force F . Its velocity at time π‘ seconds is given by the relation v i ( π‘ ) = ( 6 π π‘ + π ) / m s , where i is a unit vector in the direction of its motion, and π and π are constants. Given that the initial velocity of the body v i ο¦ = 1 5 / m s and F i = ( 1 2 π ) N , find the bodyβs speed at π‘ = 1 4 s e c o n d s .

Q16:

Three forces, β πΉ = ο» π β π + 4 β π β 9 β π ο 1 N , β πΉ = ο» 3 β π β 8 β π + π β π ο 2 N , and β πΉ = ο» 4 β π + π β π + 8 β π ο 3 N , where β π , β π , and β π are three perpendicular unit vectors, are acting upon a body of unit mass. If the displacement vector of the particle is β π = ο ( 4 π‘ ) β π + οΉ 6 π‘ + 3 π‘ ο β π + οΉ 8 π‘ + 7 ο β π ο 2 2 m , determine the constants π , π , and π .

Q17:

A body of mass 478 g has an acceleration of οΊ β 4 β π + 3 β π ο m/s^{2}, where β π and β π are perpendicular unit vectors. What is the magnitude of the force acting on the body?

Q18:

A body of mass 1 kg was moving in a straight line with a velocity β π£ = οΊ 8 β π β 8 β π ο / m s , where β π and β π are two perpendicular unit vectors. The force β πΉ = οΊ β 4 β π β 5 β π ο N acted on the body for 8 seconds. Find the bodyβs speed after the action of this force.

Q19:

A body of mass 2 kg moves in a horizontal plane in which β π and β π are perpendicular unit vectors. At time π‘ seconds ( π‘ β₯ 0 ) , the force acting on the particle is given by β πΉ = ο ( 8 π‘ β 8 ) β π + ( 4 π‘ β 3 ) β π ο N . Find the speed of the body, π£ , and its distance from the origin, π , when π‘ = 3 s .

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