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3 years ago

Assuming a constant force, if the mass of an object increases, the acceleration of the object will

Physics

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

Answer:

Assuming a constant force, if the mass of object increases, the acceleration will decrease and if mass of a object decreases, the acceleration will increase. Also if the force applied to object increases the acceleration will also increase whereas if the force applied decreases, the acceleration will also decrease.

Explanation:

We know that according to newton's second law of motion,

F = ma, where F is force applied, m is mass of an object and a is measure of body accelerated with the applies force.

a = F/m

Assuming force is constant, we get

a α 1/m, that is acceleration is inversely proportional to mass. Hence if mass of an object decreases the acceleration of object will increases and vice versa.

Also from equation, F = ma, we infer that F α a . That is the force applied is directly proportional to amount the body is accelerated. . Hence if force applied increases then the acceleration will also increase and vice versa

Anettt [7]3 years ago

4 0

For Edgeunity its decrease, increase, increase, and decrease

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The answers to the problem are as follows:

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3 years ago

A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas

Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the block

F₁ = 65 N, horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight of the block

W= m*g = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force (FN)

We apply the formula (1)

∑Fy = m*ay ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N= 0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

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