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.)
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)
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
To find the rate at which the block's acceleration is changing, we need to differentiate the acceleration with respect to time.
First, let's convert the angle from degrees to radians. We know that 1 radian is equal to π/180 degrees, so we can multiply the angle (θ) by π/180 to convert it to radians.
(a) θ(t) = (3 × 10^(-2) deg/s) t = (3 × 10^(-2) × π/180 rad/s) t
(b) θ(t) = (-3 × 10^(-2) deg/s) t = (-3 × 10^(-2) × π/180 rad/s) t
Now, let's denote the acceleration as a and the angle as θ.
The block's acceleration can be calculated using the following formula:
a = F/m = (20 N) / (7 kg) = 20/7 m/s^2
To find the rate of change of acceleration, we differentiate the acceleration with respect to time.
(a) da/dt = d(20/7 m/s^2) / dt = 0 m/s^2 (as the acceleration is not changing with time in this case)
(b) da/dt = d(20/7 m/s^2) / dt = 0 m/s^2 (as the acceleration is not changing with time in this case)
Therefore, the rate at which the block's acceleration is changing is 0 m/s^2 for both cases (a) and (b).
To find the rate at which the block's acceleration is changing, we need to determine the value of d²x/dt², where x represents the displacement of the block.
Given that the force applied to the block is 20 N and that the floor is frictionless, we can use Newton's second law, F = ma, to relate the force, mass, and acceleration of the block. However, since the angle θ is variable, we first need to resolve the applied force into its x and y components.
Let's start by finding the x-component of the applied force (F_x):
F_x = F * cos(θ)
Similarly, the y-component of the applied force (F_y) can be found as:
F_y = F * sin(θ)
Given that F = 20 N, we can substitute it into the equations:
F_x = 20 N * cos(θ)
F_y = 20 N * sin(θ)
Next, we need to relate the x-component of the applied force to the block's acceleration in the x-direction. In this case, the x-direction is along the floor, so there is no friction to oppose the motion. Thus, F_x is equal to the product of the mass (m) and the acceleration (a) in the x-direction:
F_x = ma
We can substitute the expression for F_x:
20 N * cos(θ) = ma
Rearranging the equation, we have:
a = (20 N * cos(θ)) / m
Now, since the block is being pulled along the floor, the angle θ is changing with time. We are given two different expressions for θ(t):
(a) θ(t) = (3 × 10^(-2) deg/s) * t
(b) θ(t) = (-3 × 10^(-2) deg/s) * t
To simplify the calculations, we should convert the angles into radians. Since 180 degrees is equal to π radians, we can multiply the angle in degrees by π/180 to obtain the angle in radians.
(a) θ(t) = (3 × 10^(-2) * π/180 rad/s) * t
(b) θ(t) = (-3 × 10^(-2) * π/180 rad/s) * t
Now, we can substitute the expression for θ(t) into the equation for acceleration (a) and calculate the rate at which the block's acceleration is changing by taking the second derivative of x with respect to time (t).
(a) For θ(t) = (3 × 10^(-2) * π/180 rad/s) * t:
a = (20 N * cos((3 × 10^(-2) * π/180) * t)) / m
Differentiating once with respect to time:
da/dt = (-20 N * sin((3 × 10^(-2) * π/180) * t) * (3 × 10^(-2) * π/180)) / m
(b) For θ(t) = (-3 × 10^(-2) * π/180 rad/s) * t:
a = (20 N * cos((-3 × 10^(-2) * π/180) * t)) / m
Differentiating once with respect to time:
da/dt = (-20 N * sin((-3 × 10^(-2) * π/180) * t) * (-3 × 10^(-2) * π/180)) / m
Now, you can substitute the given values for the mass (m) and calculate the rate at which the block's acceleration is changing by plugging in different values of t.