Question

A 7.00-kg object undergoes an acceleration given by ax=3.00 and ay= 9.00 m/s2. Find (a) the components of the force acting on the object and (b) the magnitude of the resultant force.

Answer 1

Answer:

(a) Fx = 21 N, Fy = 63 N

(b) 66.41 N

Explanation:

(a)

Note: the components of the force acting on the object is the force acting on the horizontal axis (Fx) as well as the vertical axis (Fy)

Fx = m(ax).............................. Equation 1

Where Fx = Horizontal force acting on the object, m = mass of the object, ax = acceleration of the object in the horizontal axis.

Given: m = 7 kg, ax = 3.0 m/s²

Substitute into equation 1

Fx = 7(3)

Fx = 21 N.

Similarly,

For the y- axis,

Fy = may.......................... Equation 2

Where Fy = vertical force acting on the object, m = mass of the object, ay = Vertical acceleration of the object

Given: m = 7 kg, ay = 9.0 m/s²

Substitute into equation 2

Fy = 7(9)

Fy = 63 N.

(b)

Assuming the horizontal and the vertical force are at right angle,

Using Pythagoras theorem,

Fr = √(Fx²+Fy²).................... Equation 3

Where Fr = magnitude of the resultant force

Given: Fx = 21 N, Fy = 63 N

Substitute into equation 3

Fr = √(21²+63²)

Fr = √(441+3969)

Fr = √(4410)

Fr = 66.41 N.

Hence the magnitude of the resultant force = 66.41 N

Answer 2

Answer:

66.4 N

Explanation:

From Newton's second law, F = ma

where F is the force, m is the mass and a is the acceleration.

Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:

$F_x = ma_x = (7.00\text{ kg})(3.00 \text{ m/s}^2) = 21.0\text{ N}$

$F_y = ma_y = (7.00\text{ kg})(9.00 \text{ m/s}^2) = 63.0\text{ N}$

The magnitude of the resultant force is given by

$F = \sqrt{F_x^2+F_y^2}$

$F = \sqrt{(21.0\text{ N})^2+(63.0\text{ N})^2} = \sqrt{(441.0\text{ N}^2)+(3969.0\text{ N}^2)} = \sqrt{(4410\text{ N}^2)} = 66.4 \text{ N}$