If motion of a particle of mass (m) is given by y=ut+1/2gt^2 (square).find force acting on it
I assume that u and g are constants (in fact initial velocity and gravity)
dy/dt = speed = u + g t
d^2y/dt^2 = acceleration = g
F = m a
so
F = m g
To find the force acting on a particle, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, we need to calculate the acceleration of the particle.
Given:
Position equation: y = ut + (1/2)gt^2
To find the acceleration, we need to take the second derivative of the position equation with respect to time (t). Let's go step by step:
Step 1: Take the first derivative of the position equation to find the particle's velocity (v):
v = dy/dt = d(ut + (1/2)gt^2)/dt
v = u + gt
Step 2: Take the second derivative of the position equation to find the particle's acceleration (a):
a = dv/dt = d(u + gt)/dt
a = g
From the above steps, we find that the acceleration (a) of the particle is equal to gravity (g). Now we can use Newton's second law to calculate the force:
F = m * a
F = m * g
Therefore, the force acting on the particle is equal to the product of the mass (m) and the acceleration due to gravity (g).
Not too bad
Wrote
dy/dt = speed = u + g t
d^2y/dt^2 = acceleration = g
F = m a
so
F = m g