If the magnitudes of the forces are F1 = 57.5 N and F2 = 37.5 N, what are the magnitude (in m/s2) and direction of the acceleration of the block? Let m = 8.00 kg and 𝜃 = 30.0°.magnitude
Question
If the magnitudes of the forces are F1 = 57.5 N and F2 = 37.5 N, what are the magnitude (in m/s2) and direction of the acceleration of the block? Let m = 8.00 kg and 𝜃 = 30.0°.magnitude
Solution
To find the acceleration, we first need to find the net force acting on the block. The forces F1 and F2 are acting at an angle θ to each other, so we can't just add them directly. We need to resolve them into their components and add those.
The components of F1 are: F1x = F1 * cos(θ) = 57.5N * cos(30) = 49.79N F1y = F1 * sin(θ) = 57.5N * sin(30) = 28.75N
The components of F2 are: F2x = F2 * cos(θ) = 37.5N * cos(30) = 32.48N F2y = F2 * sin(θ) = 37.5N * sin(30) = 18.75N
The net force in the x-direction is F1x + F2x = 49.79N + 32.48N = 82.27N The net force in the y-direction is F1y + F2y = 28.75N + 18.75N = 47.5N
The magnitude of the net force is sqrt((F_net_x)^2 + (F_net_y)^2) = sqrt((82.27N)^2 + (47.5N)^2) = 94.77N
The direction of the net force is tan^-1(F_net_y / F_net_x) = tan^-1(47.5N / 82.27N) = 30.0°
Now we can find the acceleration using Newton's second law, F = ma: a = F/m = 94.77N / 8.00kg = 11.85 m/s^2
So the magnitude of the acceleration is 11.85 m/s^2 and the direction is 30.0°.
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