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A body of mass 2.5 kg is placed on a smooth inclined plane with an angle of inclination of 65°. If the initial velocity of the body is 4 m/s, calculate its velocity after 2 seconds.

Physics

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Answer

22.1 m/s\text{22.1 m/s}

Step-by-step solution

3 steps
  1. Identify the given values.

    The mass of the body m=2.5 kgm = 2.5 \text{ kg}.
    The angle of inclination θ=65∘\theta = 65^\circ.
    The initial velocity u=4 m/su = 4 \text{ m/s}.
    The time t=2 st = 2 \text{ s}.
    The acceleration due to gravity g=10 m/s2g = 10 \text{ m/s}^2.

  2. Calculate the acceleration of the body down the smooth inclined plane.

    For a smooth inclined plane, the acceleration aa is given by the component of gravity acting along the incline.

    a=gsin⁡θa = g \sin \theta

    Substitute the values:

    a=10 m/s2×sin⁡(65∘)a = 10 \text{ m/s}^2 \times \sin(65^\circ) a=10 m/s2×0.9063a = 10 \text{ m/s}^2 \times 0.9063 a=9.063 m/s2a = 9.063 \text{ m/s}^2
  3. Use the kinematic equation to find the final velocity.

    The final velocity vv can be found using the equation:

    v=u+atv = u + at

    Substitute the initial velocity, acceleration, and time:

    v=4 m/s+(9.063 m/s2×2 s)v = 4 \text{ m/s} + (9.063 \text{ m/s}^2 \times 2 \text{ s}) v=4 m/s+18.126 m/sv = 4 \text{ m/s} + 18.126 \text{ m/s} v=22.126 m/sv = 22.126 \text{ m/s}

    Rounding to one decimal place:

    v≈22.1 m/sv \approx 22.1 \text{ m/s}

    The velocity after 2 seconds is 22.1 m/s\boxed{\text{22.1 m/s}}.

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