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Calculate acceleration (a) and final velocity (v_f).

Physics

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13. An object of mass 24 kg is accelerated up a frictionless plane inclined at an angle of 37° with the horizontal by a constant force. Starting at the bottom from the rest, it covers a distance of 18m in 3.0secs.

Given: m=24 kgm = 24 \text{ kg}, θ=37∘\theta = 37^\circ, v0=0 m/sv_0 = 0 \text{ m/s}, d=18 md = 18 \text{ m}, t=3.0 st = 3.0 \text{ s}. Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Step 1: Calculate acceleration (aa) and final velocity (vfv_f).

d=v0t+12at2  ⟹  18=0⋅3+12a(3)2d = v_0 t + \frac{1}{2} a t^2 \implies 18 = 0 \cdot 3 + \frac{1}{2} a (3)^2 18=4.5a  ⟹  a=184.5=4.0 m/s218 = 4.5a \implies a = \frac{18}{4.5} = 4.0 \text{ m/s}^2 vf=v0+at  ⟹  vf=0+(4.0)(3.0)=12 m/sv_f = v_0 + at \implies v_f = 0 + (4.0)(3.0) = 12 \text{ m/s}

Step 2: Calculate the applied force (FappF_{app}).
The net force along the incline is Fnet=maF_{net} = ma. The forces are the applied force and the component of gravity along the incline (mgsin⁡θmg\sin\theta).

Fapp−mgsin⁡θ=maF_{app} - mg\sin\theta = ma Fapp=ma+mgsin⁡θ=(24 kg)(4.0 m/s2)+(24 kg)(9.8 m/s2)sin⁡(37∘)F_{app} = ma + mg\sin\theta = (24 \text{ kg})(4.0 \text{ m/s}^2) + (24 \text{ kg})(9.8 \text{ m/s}^2)\sin(37^\circ) Fapp=96 N+235.2 N⋅0.6018≈96 N+141.56 N=237.56 NF_{app} = 96 \text{ N} + 235.2 \text{ N} \cdot 0.6018 \approx 96 \text{ N} + 141.56 \text{ N} = 237.56 \text{ N}

Part a· 1 step

What is the average power required to accomplish the process?

  1. Calculate the work done (WW) and average power (PavgP_{avg}).

    W=Fapp⋅d=(237.56 N)(18 m)=4276.08 JW = F_{app} \cdot d = (237.56 \text{ N})(18 \text{ m}) = 4276.08 \text{ J} Pavg=Wt=4276.08 J3.0 s=1425.36 WP_{avg} = \frac{W}{t} = \frac{4276.08 \text{ J}}{3.0 \text{ s}} = 1425.36 \text{ W}

    The average power is 1.43×103 W\boxed{1.43 \times 10^3 \text{ W}}.

Part b· 6 steps

What is the instantaneous power required at the end of the 3.0 second interval?

  1. Calculate the instantaneous power (PinstP_{inst}).

    Pinst=Fapp⋅vf=(237.56 N)(12 m/s)=2850.72 WP_{inst} = F_{app} \cdot v_f = (237.56 \text{ N})(12 \text{ m/s}) = 2850.72 \text{ W}

    The instantaneous power is 2.85×103 W\boxed{2.85 \times 10^3 \text{ W}}.

    14. A body is pulled along a horizontal force of 20N, calculate the work done in moving a distance of 40m.

    Given: F=20 NF = 20 \text{ N}, d=40 md = 40 \text{ m}.

  2. Calculate the work done (WW).

    W=F⋅dW = F \cdot d W=(20 N)(40 m)=800 JW = (20 \text{ N})(40 \text{ m}) = 800 \text{ J}

    The work done is 800 J\boxed{800 \text{ J}}.

    15. What is the velocity ratio of an inclined plane of length 6m if the higher end of the plane is 2m above the ground?

    Given: Length of inclined plane L=6 mL = 6 \text{ m}, Height h=2 mh = 2 \text{ m}.

  3. Calculate the velocity ratio (VR).

    VR=Length of inclineVertical heightVR = \frac{\text{Length of incline}}{\text{Vertical height}} VR=6 m2 m=3VR = \frac{6 \text{ m}}{2 \text{ m}} = 3

    The velocity ratio is 3\boxed{3}.

    16. A machine of efficiency of 70% is used to raise a body of mass 80kg through a vertical distance of 3m in 40 seconds. Calculate the power input. (Take g = 10 m/s²)

    Given: η=70%=0.70\eta = 70\% = 0.70, m=80 kgm = 80 \text{ kg}, h=3 mh = 3 \text{ m}, t=40 st = 40 \text{ s}, g=10 m/s2g = 10 \text{ m/s}^2.

  4. Calculate the work output (WoutW_{out}).

    Wout=mgh=(80 kg)(10 m/s2)(3 m)=2400 JW_{out} = mgh = (80 \text{ kg})(10 \text{ m/s}^2)(3 \text{ m}) = 2400 \text{ J}

  5. Calculate the power output (PoutP_{out}).

    Pout=Woutt=2400 J40 s=60 WP_{out} = \frac{W_{out}}{t} = \frac{2400 \text{ J}}{40 \text{ s}} = 60 \text{ W}

  6. Calculate the power input (PinP_{in}).

    η=PoutPin  ⟹  Pin=Poutη\eta = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta} Pin=60 W0.70≈85.71 WP_{in} = \frac{60 \text{ W}}{0.70} \approx 85.71 \text{ W}

    The power input is 85.7 W\boxed{85.7 \text{ W}}.

    17. (i) Give an example of a machine where the load is between the fulcrum and the effort. (ii) A wheel and axle is used to raise a man whose weight is 800N by the application of 250N. The radii of the wheel and axle are 800mm and 200mm respectively, what is the efficiency of the machine?

Part b(i)

Example of a machine where the load is between the fulcrum and the effort.

This describes a Class 2 lever.
Example: A wheelbarrow.

Part b(ii)· 3 steps

Efficiency of a wheel and axle.

Given: Load FL=800 NF_L = 800 \text{ N}, Effort FE=250 NF_E = 250 \text{ N}, Wheel radius RW=800 mm=0.8 mR_W = 800 \text{ mm} = 0.8 \text{ m}, Axle radius RA=200 mm=0.2 mR_A = 200 \text{ mm} = 0.2 \text{ m}.

  1. Calculate the Mechanical Advantage (MA).

    MA=FLFE=800 N250 N=3.2MA = \frac{F_L}{F_E} = \frac{800 \text{ N}}{250 \text{ N}} = 3.2

  2. Calculate the Velocity Ratio (VR).

    VR=RWRA=0.8 m0.2 m=4VR = \frac{R_W}{R_A} = \frac{0.8 \text{ m}}{0.2 \text{ m}} = 4

  3. Calculate the efficiency (η\eta).

    η=MAVR×100%\eta = \frac{MA}{VR} \times 100\% η=3.24×100%=0.8×100%=80%\eta = \frac{3.2}{4} \times 100\% = 0.8 \times 100\% = 80\%

    The efficiency of the machine is 80%\boxed{80\%}.

    18. A wheel and axle have radii 80cm and 10 cm respectively. If the efficiency of the machine is 0.85, what load will an applied force of 1200N to the wheel raise?

    Given: Wheel radius RW=80 cm=0.8 mR_W = 80 \text{ cm} = 0.8 \text{ m}, Axle radius $R_A =

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