An electric motor is used to pull a 125 kg box across a floor using a long cable. the tension in the cable is 350N and the box accelerates at 1.2m/s^2[forward] for 5.0s. The cable breaks and the box slows down and stops.
b) How far does the box travel up to the moment the cable breaks?
c)How far does the box travel from the moment the cable breaks until it stops?
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a) (1/2) a t^2, where
a = 1.2 m/s^2
t = 5.0 s
b) Speed at cable breakage = a t
V' = 6 m/s
Let distance travelled after cable breakage be X.
M g X *Uk = (M/2) V'^2
You need the coefficient of kinetic friction Uk. You can get it from the initial cable tension, before it breaks.
350 - M*g*Uk = M*a
Uk = (350 - M a) /(Mg)
= 0.163
Now solve for X.
To calculate the distance the box travels up to the moment the cable breaks, we can use the kinematic equation:
d = v₀t + 0.5at²
Where:
d = distance traveled
v₀ = initial velocity (0 m/s as the box is at rest)
t = time (5.0 s)
a = acceleration (1.2 m/s² [forward])
Plugging in the values:
d = 0(5.0) + 0.5(1.2)(5.0)²
d = 0 + 0.5(1.2)(25)
d = 0 + 0.5(30)
d = 0 + 15
d = 15 meters
Therefore, the box travels 15 meters up to the moment the cable breaks.
To calculate the distance the box travels from the moment the cable breaks until it stops, we can use the kinematic equation again. However, since the box is slowing down, the acceleration will be negative:
d = v₀t + 0.5at²
Where:
d = distance traveled
v₀ = initial velocity (the velocity of the box when the cable breaks)
t = time (unknown)
a = acceleration (unknown, but it can be calculated using the final velocity)
First, let's calculate the final velocity:
v = v₀ + at
Where:
v = final velocity (0 m/s as the box stops)
v₀ = initial velocity (the velocity of the box when the cable breaks)
a = acceleration (unknown, but given as 1.2 m/s² [forward])
t = time (unknown)
Plugging in the values:
0 = v₀ + (1.2)(t)
v₀ = -1.2t
Now, let's substitute this value for v₀ in the distance equation:
d = (-1.2t)(t) + 0.5(-1.2)(t)²
d = -1.2t² - 0.6t²
d = -1.8t²
To find the time it takes for the box to stop, we need to solve for t when v = 0:
0 = -1.2t
t = 0
Therefore, the box stops at t = 0 seconds.
Now, let's calculate the distance traveled:
d = -1.8(0)²
d = 0
Therefore, the box travels 0 meters from the moment the cable breaks until it stops.
To find the distance the box travels up to the moment the cable breaks (question b), we need to calculate the distance covered during the acceleration phase. We can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Given:
mass (m) = 125 kg
tension (T) = 350 N
acceleration (a1) = 1.2 m/s^2
time (t1) = 5.0 s
First, let's find the net force acting on the box during the acceleration phase using Newton's second law:
net force (F_net) = mass (m) * acceleration (a1)
F_net = 125 kg * 1.2 m/s^2 = 150 N
Since the tension in the cable is acting in the opposite direction to the net force, we have:
Tension (T) = -F_net
T = -150 N
Now, let's calculate the initial velocity (v1):
Tension (T) = mass (m) * acceleration (a1) + mass (m) * initial velocity (v1)
-150 N = 125 kg * 1.2 m/s^2 + 125 kg * v1
Simplifying the equation:
-150 N = 150 kg·m/s + 125 kg·v1
v1 = (-150 - 150 kg·m/s) / 125 kg
v1 = -3 m/s
Using the equation mentioned earlier, we can calculate the distance (d1) covered during the acceleration phase:
d1 = (v1 * t1) + (0.5 * a1 * t1^2)
d1 = (-3 m/s * 5 s) + (0.5 * 1.2 m/s^2 * (5 s)^2)
d1 = -15 m + 15 m = 0 m
Hence, the box does not cover any distance up to the moment the cable breaks (question b).
To find the distance the box travels from the moment the cable breaks until it stops (question c), we need to calculate the distance covered during the deceleration phase.
Since the cable breaks, there is no tension force acting anymore, and the net force acting on the box during the deceleration phase is due to the friction between the box and the floor. This force opposes the motion and causes the box to slow down until it stops.
To find the deceleration (a2), we can use Newton's second law:
F_net = mass (m) * acceleration (a2)
Frictional force (F_f) = mass (m) * acceleration (a2)
The frictional force can be calculated using the normal force (F_n) and the coefficient of friction (μ), where F_n = mass (m) * gravity (g) and g ≈ 9.8 m/s^2:
Frictional force (F_f) = μ * F_n
Since the box is on a horizontal floor, the normal force is equal to the weight of the box:
F_n = m * g
Substituting the value of F_n into the frictional force equation:
Frictional force (F_f) = μ * m * g
Hence, F_net = μ * m * g, which is equal to the mass (m) times the deceleration (a2):
F_net = μ * m * g = m * a2
Now we can find the deceleration (a2):
a2 = μ * g
a2 = μ * 9.8 m/s^2
The box slows down with this deceleration until it eventually stops. To find the distance (d2) covered during this deceleration, we can use the formula:
distance (d2) = (final velocity * time) + (0.5 * deceleration * time^2)
Since the final velocity is 0 m/s (as the box stops), we have:
d2 = 0.5 * a2 * t2^2
To find the time it takes for the box to stop (t2), we can use the formula:
final velocity = initial velocity + (acceleration * time)
0 m/s = -3 m/s + (a2 * t2)
Since the initial velocity is -3 m/s:
t2 = 3 m/s / a2
Finally, we can substitute the value of t2 into the distance formula to find d2:
d2 = 0.5 * a2 * (3 m/s / a2)^2
d2 = 0.5 * a2 * 9 m^2/s^2 / a2^2
d2 = 4.5 m / a2
Therefore, the box travels a distance of 4.5 meters from the moment the cable breaks until it stops (question c).