Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Work Power and Energy question

2023 · 11 Apr · Shift 2 · Q67
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Work Power and Energy
  5. /2023 · 11 Apr · Shift 2 · Q67

Work Power and Energy question

2023 · 11 Apr · Shift 2 · Q67

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1
A block of mass 5 kg5 \mathrm{~kg}5 kg starting from rest pulled up on a smooth incline plane making an angle of 30∘30^{\circ}30∘ with horizontal with an affective acceleration of 1 ms−21 \mathrm{~ms}^{-2}1 ms−2. The power delivered by the pulling force at t=10 st=10 \mathrm{~s}t=10 s from the start is ‾\underline{\hspace{2cm}}​ W. [use g=10 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2}g=10 ms−2 ] (calculate the nearest integer value)
Numerical answer
View written solutionFree

Correct answer: 300

  1. Given data

    • Mass of block: m=5 kgm=5\,\text{kg}m=5kg
    • Inclination angle: θ=30∘\theta=30^\circθ=30∘
    • Acceleration up the plane: a=1 m s−2a=1\,\text{m s}^{-2}a=1m s−2
    • Time: t=10 st=10\,\text{s}t=10s
    • Initial velocity: u=0u=0u=0
    • Gravity: g=10 m s−2g=10\,\text{m s}^{-2}g=10m s−2

  2. Find the pulling force

    Since the plane is smooth, the forces along the incline are:

    • Pulling force FFF upward
    • Component of weight downward: mgsin⁡θmg\sin\thetamgsinθ

    Applying Newton's second law along the incline: F−mgsin⁡θ=maF - mg\sin\theta = maF−mgsinθ=ma

    Now, mgsin⁡θ=5×10×sin⁡30∘=50×12=25 Nmg\sin\theta = 5\times 10 \times \sin 30^\circ = 50\times \frac{1}{2}=25\,\text{N}mgsinθ=5×10×sin30∘=50×21​=25N

    Also, ma=5×1=5 Nma=5\times 1=5\,\text{N}ma=5×1=5N

    Therefore, F=25+5=30 NF=25+5=30\,\text{N}F=25+5=30N

  3. Find the speed at t=10 st=10\,\text{s}t=10s

    Using v=u+atv=u+atv=u+at v=0+(1)(10)=10 m s−1v=0+(1)(10)=10\,\text{m s}^{-1}v=0+(1)(10)=10m s−1

  4. Calculate power delivered by the pulling force

    Instantaneous power is: P=FvP=FvP=Fv since force and velocity are along the same direction.

    Thus, P=30×10=300 WP=30\times 10=300\,\text{W}P=30×10=300W

  5. Nearest integer value 300\boxed{300}300​

PreviousNext

More from Work Power and Energy

  • Given below are two statements: Statement I : A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provide equal retarding forces. Both come to rest in equal distance. Statement II : A car moving…2023 · MCQ
  • To maintain a speed of 80 km/h by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be ​ KJ. [The coefficient of friction between tyre of bus and road is…2023 · Numerical
  • The ratio of powers of two motors is x​+13x​​, that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value…2023 · MCQ
  • Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses respectively is :2023 · MCQ
  • A car accelerates from rest to u m/s. The energy spent in this process is E J. The energy required to accelerate the car from u m/s to 2um/s is nE J.…2023 · Numerical
  • A block of mass 10 kg is moving along x-axis under the action of force F=5x N. The work done by the force in moving the block from x=2m to 4m will be ​ J.2023 · Numerical
  • A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of 5th second. The force acted on the body is ​ N.2023 · Numerical
  • A body of mass 1kg begins to move under the action of a time dependent force F=(ti+3t2j​) N, where i and j​ are the unit vectors along x and y axis. The…2023 · Numerical