A force of 10 N acts on a mass of 2 kg for a time of 5 s. What is the distance travelled by the mass in this time given that it has an initial velocity of 5 m s^-1 in the direction opposite to that of the force?
First we"ll find the acceleration of the mass: By Newton's Second Law of Motion: F=ma a=F/m a=10/2 a=5m/s^2 Now: For distance covered by body: By 2nd equation of motion: S=vit+1/2 at^2 S=5.5+1/2 5(5)^2 S=175/2 or S=87.5m
The answer given was 43 m...
@tukitw check your question values are right
take initial velocity as -5m/s as it is in the direction opposite to the force and we have the acceleration=5m/s^2 so if we put these values in s=ut+1/2 x at^2, we get s=-5 x 5 + 5 x 5 x 5/2 therefore we get s=37.5m and this is not the answer for our question as we are asked for the distance travelled so what is exactly hapening here??? the thing happening here is the change in the direction of motion first the body was moving with some velocity and then it gained some acceleration in the oposite direction so there must be a point where the velocity had become zero and the direction of motion has been changed (try to imagine this and you will understand it more accurately) so let us find that when is this instant of change in the direction of otion is taking place by the equation v=u+at so at this instant , the velocity = 0 so here v=0 therefore, 0=-5+5t so we get the time = 1 sec when the direction of the motion is changing so the displacement during this time is s=-5 x 1 +(5 x 1 x 1)/2=-2.5 but we will take it positive(think why?) next. let us find the distance travelled after this change in the direction of motion here, t= 4 sec as one second is over and initial velocity=0 (think why) so s= 0+ (5 x 4 x 4)/2=40 so total distance travelled by the block is 40+2.5=42.5(nearly equal to 43) if you have any problem with this question , feel free to ask
thanks for the medal @Imtiaz7 and @tukitw
no problem @rajat97 yor were right i missed the statement that the force is in opposite direction
and also 'distance travelled'
thanks for helping
: )
its my pleasure
Join our real-time social learning platform and learn together with your friends!