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Math  /  Calculus

Question41. (II) A mass mm is at rest on a horizontal frictionless surface at t=0t=0. Then a constant force F0F_{0} acts on it for a time t0t_{0}. Suddenly the force doubles to 2F02 F_{0} and remains constant until t=2t0t=2 t_{0}. Determine the total distance traveled from t=0t=0 to t=2t0t=2 t_{0}.

Studdy Solution

STEP 1

1. The mass m m is initially at rest.
2. The surface is horizontal and frictionless.
3. A constant force F0 F_0 acts on the mass for a time t0 t_0 .
4. The force doubles to 2F0 2F_0 and acts until time t=2t0 t = 2t_0 .

STEP 2

1. Calculate the acceleration during each time interval.
2. Determine the velocity at the end of the first interval.
3. Calculate the distance traveled during each interval.
4. Sum the distances to find the total distance traveled.

STEP 3

Calculate the acceleration during each time interval.
For the first interval 0≤t≤t0 0 \leq t \leq t_0 : a1=F0m a_1 = \frac{F_0}{m}
For the second interval t0<t≤2t0 t_0 < t \leq 2t_0 : a2=2F0m a_2 = \frac{2F_0}{m}

STEP 4

Determine the velocity at the end of the first interval.
Using the formula v=u+at v = u + at , where u=0 u = 0 (initial velocity): v1=0+a1⋅t0=F0m⋅t0 v_1 = 0 + a_1 \cdot t_0 = \frac{F_0}{m} \cdot t_0

STEP 5

Calculate the distance traveled during each interval.
For the first interval 0≤t≤t0 0 \leq t \leq t_0 : Using s=ut+12at2 s = ut + \frac{1}{2}at^2 : s1=0⋅t0+12⋅F0m⋅t02=12⋅F0m⋅t02 s_1 = 0 \cdot t_0 + \frac{1}{2} \cdot \frac{F_0}{m} \cdot t_0^2 = \frac{1}{2} \cdot \frac{F_0}{m} \cdot t_0^2
For the second interval t0<t≤2t0 t_0 < t \leq 2t_0 : Initial velocity for this interval is v1 v_1 : s2=v1⋅t0+12⋅a2⋅t02 s_2 = v_1 \cdot t_0 + \frac{1}{2} \cdot a_2 \cdot t_0^2 =(F0m⋅t0)⋅t0+12⋅2F0m⋅t02 = \left(\frac{F_0}{m} \cdot t_0\right) \cdot t_0 + \frac{1}{2} \cdot \frac{2F_0}{m} \cdot t_0^2 =F0m⋅t02+F0m⋅t02 = \frac{F_0}{m} \cdot t_0^2 + \frac{F_0}{m} \cdot t_0^2 =2⋅F0m⋅t02 = 2 \cdot \frac{F_0}{m} \cdot t_0^2

STEP 6

Sum the distances to find the total distance traveled.
stotal=s1+s2 s_{\text{total}} = s_1 + s_2 =12⋅F0m⋅t02+2⋅F0m⋅t02 = \frac{1}{2} \cdot \frac{F_0}{m} \cdot t_0^2 + 2 \cdot \frac{F_0}{m} \cdot t_0^2 =12⋅F0m⋅t02+42⋅F0m⋅t02 = \frac{1}{2} \cdot \frac{F_0}{m} \cdot t_0^2 + \frac{4}{2} \cdot \frac{F_0}{m} \cdot t_0^2 =52⋅F0m⋅t02 = \frac{5}{2} \cdot \frac{F_0}{m} \cdot t_0^2
The total distance traveled from t=0 t=0 to t=2t0 t=2t_0 is:
52⋅F0m⋅t02 \boxed{\frac{5}{2} \cdot \frac{F_0}{m} \cdot t_0^2}

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