Home > Physics > Newton's Second Law

An object with a mass of #3 kg# is acted on by two forces. The first is #F_1= < 1 N , 4 N># and the second is #F_2 = < 2 N, -3 N>#. What is the object's rate and direction of acceleration?

Answer 1

Charlotte Aaron

Rate of acceleration
#abs (vec a) = sqrt(1^2 + (1/3)^2) = sqrt(10)/3 \ m/s^2#

Direction
#arctan (1/3)# measured CCW from x-axis

3 ((a_x), (a_y))# #((3),(1))

((1),(1/3))# #((a_x), (a_y))

#abs (vec a) = sqrt(1^2 + (1/3)^2) = sqrt(10)/3 \ m/s^2# is the acceleration rate.

#arctan (1/3)#, measured CCW from the x-axis, is the direction.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Xavier Adame

To find the object's rate and direction of acceleration, calculate the net force by summing the individual forces and then use Newton's second law (F = ma) where the acceleration (a) equals the net force divided by the mass (m). The net force (F_net) is the vector sum of F_1 and F_2: F_net = F_1 + F_2 = <1 N + 2 N, 4 N + (-3 N)> = <3 N, 1 N> Given that the mass (m) is 3 kg, use Newton's second law to find the acceleration (a): a = F_net / m = <3 N / 3 kg, 1 N / 3 kg> = <1 m/s^2, 1/3 m/s^2> So, the object's rate of acceleration is 1 m/s^2 in the x-direction and 1/3 m/s^2 in the y-direction.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

An object with a mass of #3 kg# is acted on by two forces. The first is #F_1= &lt; 1 N , 4 N&gt;# and the second is #F_2 = &lt; 2 N, -3 N&gt;#. What is the object's rate and direction of acceleration?

An object with a mass of #3 kg# is acted on by two forces. The first is #F_1= < 1 N , 4 N># and the second is #F_2 = < 2 N, -3 N>#. What is the object's rate and direction of acceleration?