We can do this by taking the square root of the sum The question has asked us to find the magnitude of theįorce, so we will need to find the magnitude of this vector. This is the vector form of the force acting on the body. Substituting in our values for 𝑚 and ⃑ 𝑎, we obtain So the formula we can use to find the force acting on the body is Now, we can see that our acceleration is given as a vector, In doing so, we can say that the mass of the body isĠ.478 kg. Terms of N, we will need to convert the mass to What is the magnitude of theįirst, let us note the units that are used in the question. Let us consider Newton’s second law of motion when it is applied to a system where the acceleration is given inĮxample 3: Finding the Magnitude of a Force Acting on a Body given the Acceleration Vector and MassĪ body of mass 478 g has an acceleration ofĪnd ⃑ 𝑗 are perpendicular unit vectors. The force acting on the body has a magnitude given by
To determine the acceleration, the formula must be rearranged to make The value of 𝑎 will be negative, as the force acts in the opposite direction to 𝑣. 𝑢 is the speed of the body at the beginning of the interval, and 𝑠 is theĭisplacement of the body along the line ofĪction of the force that acts on it during the interval. Where 𝑣 is the speed of the body at the end of the interval, Traveled is stated in the question, we can calculate the acceleration using the The acceleration of the body is required. That caused this change in the body’s motion. As a result, over the nextĢ6 m, its speed decreased uniformly to 12 m/s. Started acting on the body, opposing its motion. Let us look at such an example.Įxample 2: Finding the Magnitude of the Force Acting on a Moving Body in the Opposite Direction to Reduce Its VelocityĪ body of mass 41 kg was moving along a horizontal road at 14 m/s. The force that acts to produce this acceleration acts on a mass of 9 kg,Īnd therefore the magnitude of the force is given byĪ net force acting in the opposite direction to the direction in which an object is moving acts to decrease the velocity of the object. The increase in speed occurs in a time interval of 1 2 a second, The increase in speed, Δ 𝑣, is therefore given by 𝑣 = 5 8 × 1 0 0 0 3 6 0 0 = 1 4 5 9 /, i n i t i a l m sĪnd 66 km/h in metres per second is given by Question to speeds in metres per second ( m/s).
Of metres per second squared ( m/s 2), so it will be necessary to convert the speeds in the Acceleration is usually expressed in units The acceleration of the particle is required. Example 1: Finding a Constant Force Using Newton’s Second Law of Motionġ 2 a second.
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