Posted by **John** on Tuesday, January 21, 2014 at 3:05pm.

The figure shows a 7 kg block being pulled along a frictionless floor by a cord that applies a force of constant magnitude 20 N but with an angle θ(t) that varies with time. When angle θ = 29°, at what rate is the block's acceleration changing if (a)θ(t) = (3 × 10-2 deg/s)t and (b)θ(t) = (-3 × 10-2 deg/s)t ? (Hint: Switch to radians.)

- Physics -
**Damon**, Tuesday, January 21, 2014 at 3:33pm
I assume that Theta is up and down from horizontal

calling theta = A

d sin A /dA = cos A

d sin A/dt = d sin A/dA * dA/dt

so

d sin A /dt = cos A (dA/dt )

similarly

d cos A/dt = -sin A dA/dt

That settled, let's look at the problem

Force = m * a

a = F/7

da/dt = (1/7) dF/dt

F in direction of motion = 20 cos A

d F/dt =20 d/dt(cosA)

but we already know what that is

dF/dt = 20 (-sin A) dA/dt

A = 28 deg = .489 radian

sin A = .469

dA/dt = 3*10^-2 (pi/180) = 5.24*10^-5 rads/s

so

dF/dt = 20 (-.489)(5.24*10^-5)

da/dt =(1/7)dF/dt = (20/7)(-.489)(5.24*10^-5)

- Physics -
**bobpursley**, Tuesday, January 21, 2014 at 3:34pm
Net force=mass*acceleration

20*cosTheta=mass* acceleratin

take the deriviative...

-20SinTheta dTheta/dt=mass*d(acceleration)/dt

yes, dTheta/dt is in radians/sec, so convert it. solve for d(acceleration)/dt and the units will be m/sec^3

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