An object with a mass of $5 kg$ $5"kg"$ is hanging from a spring with a constant of $3 {kg/s}^{2}$ $3 "kg/s"^2$ . If the spring is stretched by $3 m$ $3"m"$ , what is the net force on the object?

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Answer 1 · 2016-11-09T07:11:39

$41 N$ $41 "N"$ downwards.

The gravitational force acting on the object is

$W = m g = (5 kg) \cdot (10 {m/s}^{2}) = 50 N$ $W = mg = (5 "kg") * (10 "m/s"^2) = 50 "N"$

in the downwards direction.

The spring force is given by Hooke's Law,

$F_{Spring} = k x = (3 {kg/s}^{2}) \cdot (3 m) = 9 N$ $F_"Spring" = kx = (3 "kg/s"^2) * (3 "m") = 9 "N"$

in the upwards direction.

Therefore, the net force is the sum of all forces acting on the object.

$F_{Net} = W - F_{Spring}$ $F_"Net" = W - F_"Spring"$

$= 50 N - 9 N$ $= 50 "N" - 9 "N"$

$= 41 N$ $= 41 "N"$

in the downwards direction.

An object with a mass of 5"kg"5kg5"kg" is hanging from a spring with a constant of 3 "kg/s"^23kg/s23 "kg/s"^2. If the spring is stretched by 3"m"3m3"m", what is the net force on the object?