A body of mass 20 kg is moving with a speed of 5 m/s. The distance travelled by the body before coming to rest when a constant retarding force of 40 N is applied on it.
Answers
Answer:
6s
Explanation:
From Newton's IInd law of motion F
net
net
net =ma
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a=
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+at
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s;
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2 ⇒t=
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2 ⇒t= 2.5
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2 ⇒t= 2.515
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2 ⇒t= 2.515
net =maFrom given data in question F=−50N;m=20kg;u=15m/s;v=0m/s⇒a= 20−50 =−2.5m/s 2 Using kinematics equation v=u+atwhere v and u are final and initial velocities and a is acceleration and t is time in s; v=0;u=15m/s;a=−2.5m/s 2 ⇒t= 2.515 =6s
HOPE THAT THIS WILL HELP YOU MATE.
Answer:
7.5 m
Explanation:
if mass is 20 kg and force 40N then from
F=ma
a=F/m=40/20=2
now,
a=∆v/∆t
∆t=∆v/a=5/2=2.5 s
S=ut + 1/2(at^2)= 5×2.5+ 6.25=1.25+6.25=7.5